Subtopic Archive - Page 9 of 10

Applications of exponential functions (Intensity of earthquakes and sound loudness)

August 12, 2019/by admin_ct

Intensity of earthquakes and sound loudness

Richter scale is a logarithmic function to measure strength of earthquakes and its formula is given as,

$\dpi{150} \mathbf{R=log \left ( \frac{I}{I_{0}} \right )}$

Here R is the Richter scale magnitude of earthquake and $\dpi{120} \frac{I}{I_{0}}$ is the intensity ratio of earthquakes.

The intensity of earthquake will typically measure between 2 and 10 on Richter scale.

Example1: An earthquake is measured with a wave intensity 200 times as great as I₀. What is the magnitude of this earthquake using the Richter scale, to the nearest tenth?

Solution: Given that wave intensity 200 times as great as I₀

So $\dpi{120} I= 200I_{0}$

Substituting these values into the formula, we get

R = log(200) = 2.3

Magnitude of this earthquake is 2.3 on Richter scale.

Above formula can also be used in another form when Richter magnitude of two earthquakes is given and we need to compare their intensities.

$\dpi{150} \mathbf{\frac{I}{I_{0}}= 10^{M-M_{0}}}$

Where $\dpi{120} M_{0}$ is the magnitude of weaker earthquake and M is the magnitude of stronger earthquake.

Example2: A small tremor of magnitude 3.4 is followed by strong one of magnitude 4.1. How much stronger is the second tremor than the first one.

Solution: Here we need to find intensity of earthquakes.

Using formula $\dpi{120} \frac{I}{I_{0}}= 10^{M-M_{0}}$

$\dpi{120} \frac{I}{I_{0}}= 10^{4.1-3.4}$

$\dpi{120} \frac{I}{I_{0}}= 5.01$

Second tremor is 5 times stronger than the first one.

Example3: A strong earthquake with a magnitude of 5.7 is three times more intense than the weaker earthquake. Find the magnitude of weaker earthquake

Solution: Given that M =5.7 and $\dpi{120} I=3I_{0}$ . We need to find $\dpi{120} M_{0}$ .

$\dpi{120} \frac{3I_{0}}{I_{0}}=10^{5.7-M_{0}}$

$\dpi{120} 3=10^{5.7-M_{0}}$

$\dpi{120} log3=log10^{5.7-M_{0}}$

$\dpi{120} log3=(5.7-M_{0})log10$

$\dpi{120} 0.477 = 5.7-M_{0}$

$\dpi{120} M_{0}=5.02$

Formula for loudness of sound using sound intensity is :

$\dpi{150} \mathbf{dB-dB_{0}= 10log\left ( \frac{I}{I_{0}} \right )}$

Where dB is decibel value of loud sound and $\dpi{120} dB_{0}$ is decibel value of soft sound. $\dpi{120} \left ( \frac{I}{I_{0}} \right )$ is the intensity ratio of sounds.

Example4: A sound of 75dB is 50 times louder than a weaker sound. What is the loudness of weaker sound.

Solution: $dB-dB_{0}= 10log\left ( \frac{I}{I_{0}} \right )$

$75-dB_{0}= 10log\left ( \frac{50I_{0}}{I_{0}} \right )$

$75-dB_{0}= 10log\left ( \50 \right )$

$75-dB_{0}= 16.989$

$dB_{0}= 75-16.99= 58.01$

Example5: Two different sounds has their loudness level as 80 dB and 100dB. How much louder is the second sound?

Solution: $dB-dB_{0}= 10log\left ( \frac{I}{I_{0}} \right )$

$100-80= 10log\left ( \frac{I}{I_{0}} \right )$

$\frac{20}{10}= log\left ( \frac{I}{I_{0}} \right )$

$2= log\left ( \frac{I}{I_{0}} \right )$

$10^{2}=\frac{I}{I_{0}}$

$100 I_{0}=I$

So the louder sound is 100 times intense than the weaker sound.

Formula to find acid strength is:

$\dpi{150} \mathbf{pH=-log[H^{+}]}$

Where is the concentration of hydrogen ion and pH is acid strength(power of hydrogen)

Example6: Calculate the pH of the acid which has concentration of $\dpi{120} {\color{Red} H^{+}}$ ions as $\dpi{120} {\color{Red} 2.5\times 10^{-5}}$ .

Solution: $pH=-log[H^{+}]$

$pH=-log[2.5\times 10^{-5}]$

pH = 4.60

Example7: Find hydrogen concentration of acid which has a pH value of 3.7

Solution: $pH=-log[H^{+}]$

$3.7=-log[H^{+}]$

$-3.7=log[H^{+}]$

$10^{-3.7}=[H^{+}]$

$\dpi{120} 1.99\times 10^{-4} mol/L=\left [ H^{+} \right ]$

Practice problems:

A small tremor of magnitude 2.3 is followed by strong one of magnitude 5.3. How much stronger is the second tremor than the first one.
A strong earthquake with a magnitude of 5.9 is one thousand times more intense than the weaker earthquake. Find the magnitude of weaker earthquake.
A sound of 70dB is 8900 times louder than a weaker sound. What is the loudness of weaker sound.
Two different sounds has their loudness level as 97 dB and 110dB. How much louder is the second sound?
Find hydrogen concentration of acid which has a pH value of 8.7

Answers:

1000
2.9
30.5 dB
20
$\dpi{120} 2\times 10^{-9}$ mol/L

Applications of exponential and logarithmic functions (Population and bacteria growth)

August 11, 2019/by admin_ct

Population and bacteria growth

Exponential growth and decay: For growth and decay word problems we use formula:

$\dpi{150} \mathbf{A=A_{0}\left ( b \right )^{\frac{t}{p}}}$

Where A=Amount after certain time

$A_{0}$ =Initial amount

b = Type of growth

t = time elapsed

p= period for growth to occur

Example1: A bacterial culture doubles every 3 hours. If the culture started with 20,000 bacteria, how many will be present in 4 hrs.

Solution: Given that culture doubles every 3 hrs.

That means b=2 and p= 3

Also given initial amount $A_{0}= 20,000$ and time t= 4

Using formula, $\dpi{120} A=A_{0}\left ( b \right )^{\frac{t}{p}}$

$A= 20,000\left ( 2\right )^{\frac{4}{3}}$

A = 50396.8419

A = 50397 bacteria

Example 2: Population of a town triples every 5 years. If 4000 people were present in year 2005, then find the population in year 2015.

Solution: Given that population triples every 5 years.

That means b=3 and p= 5

Also given initial population $A_{0}= 4,000$

Time elapsed (t)=2015-2005= 10 years.

Using all these values into same formula, we get

$A= 4,000\left ( 3\right )^{\frac{10}{5}}$

A = 4000(9) =36,000

Solving for variable other than A.

Example3: Population of a town changes by exponential growth factor b every 4 years. If 2350 grows to 7000 in 3 years, what is the value of b.

Solution: Given initial population $A_{0}= 2350$

Final population A =7000

Time elapsed (t)= 3 yrs

p= 4 yrs

using formula, $\dpi{120} A=A_{0}\left ( b \right )^{\frac{t}{p}}$

$7000= 2350\left ( b\right )^{\frac{3}{4}}$

$\frac{7000}{2350}= \left ( b\right )^{\frac{3}{4}}$

$\ln \left (\frac{7000}{2350} \right )= \ln \left ( b\right )^{\frac{3}{4}}$ (taking ln of both sides)

$\ln \left (\frac{7000}{2350} \right )= \frac{3}{4}\ln \left ( b\right )$ (using formula $\dpi{120} \ln x^{y}= y\ln x$ )

$\frac{4}{3}\ln \left (\frac{140}{47} \right )= \ln \left ( b\right )$
1.4553.. = ln(b)

$\dpi{120} e^{1.4553...}= b$

4.3 = b

Example4: A bacterial culture double every 5 hours. How long will it take for a culture to quadruple?

Solution: Given that culture doubles every 5hrs.

So b=2 and p= 5

Also given, $\dpi{120} A=4A_{0}$

Using formula we get, $\dpi{120} 4A_{0}=A_{0}\left ( 2 \right )^{\frac{t}{5}}$

$\dpi{120} 4= 2^{\frac{t}{5}}$

$\dpi{120} ln4= ln2^{\frac{t}{5}}$

$\dpi{120} ln4= \frac{t}{5}ln2$

$\dpi{120} \frac{ln4}{ln2}= \frac{t}{5}$

$\dpi{120} \frac{2ln2}{ln2}= \frac{t}{5}$

$\dpi{120} 2= \frac{t}{5}$

2(5) =t => t= 10 hours

Example5: A radioactive sample has half life of 3 days. How long will it take for only 1/8^th of the sample to remain?

Solution: Give that half life of 3 days.

That means b= ½ and p= 3

Also given that A = 1/8^th of initial sample, so $\dpi{120} A= \frac{1}{8}A_{0}$

$\dpi{120} \frac{1}{8}A_{0}= A_{0}\left ( \frac{1}{2} \right )^{\frac{t}{3}}$

$\dpi{120} \frac{1}{8}= \left (\frac{1}{2} \right )^{\frac{t}{3}}$

$\dpi{120} \left ( \frac{1}{2} \right )^{3}= \left ( \frac{1}{2} \right )^{\frac{t}{3}}$ (When bases are same, powers can be equated)

3 = t/3 => t= 9 days.

Example6: Light passing through murky water retain 3/4^th of its intensity for every meter of water. At what depth will the light intensity be 60% of what it is at the surface?

Solution: Final amount is always the remaining amount. Here remaining light intensity is 60% so we can use final amount A as $\dpi{120} 0.6A_{0}$

Given that water retain3/4^th of its intensity, so b=3/4 and p=1.

Using the values into the formula we get,

$\dpi{120} 0.6A_{0}=A_{0}\left ( \frac{3}{4} \right )^{\frac{t}{1}}$

$\dpi{120} 0.6 =(0.75)^{t}$

ln(0.6) = ln(0.75)^t

ln(0.6) = t ln(0.75)

$\dpi{120} \frac{ln(0.6)}{ln(0.75)}=t$

1.78 m = t

Example7: A radioactive sample has a half life of 3 years and has initial mass of 68 g. How long will it take for the sample to loose 8g ?

Solution: Here the final amount A, would be the remaining amount . So = 68-8= 60 g.

Note : Avoid using lost amount 8g as remaining final amount.

Given half life of 3 years, that means b=1/2 and p=3

$\dpi{120} A=A_{0}\left ( b \right )^{\frac{t}{p}}$

$\dpi{120} 60= 68\left ( \frac{1}{2} \right )^{\frac{t}{3}}$

$\dpi{120} \frac{60}{68}= \left ( \frac{1}{2} \right )^{\frac{t}{3}}$

$\dpi{120} ln\frac{15}{17}= ln\left ( \frac{1}{2} \right )^{\frac{t}{3}}$

$\dpi{120} ln\frac{15}{17}= \frac{t}{3}ln\left ( \frac{1}{2} \right )$

$\dpi{120} \frac{ln\frac{15}{17}}{ln\frac{1}{2}}=\frac{t}{3}$

$\dpi{120} 3\frac{ln\frac{15}{17}}{ln\frac{1}{2}}=t$
0.54 years = t

Rate problems : For increasing rate we always add 1 to the b value and for decreasing rate we always subtract b value from 1. For these problems p is always 1.

Example8: A crack in a window grows by 4.5 % every hour. If the crack start at a length of 3 cm, how long will it be in 3 hours.

Solution: Given that Initial amount = 3 cm

And rate =4.5%= 0.045

So b =1+0.045 =1.045

Using formula, $\dpi{120} A=A_{0}\left ( b \right )^{\frac{t}{p}}$

$\dpi{120} A=3\left ( 1.045 \right )^{\frac{3}{1}}$

A = 3.42 cm

Example9: Light passing through murky water loses 30% of its intensity for every meter of water. At what depth will the intensity be half of what it is at the surface?

Solution: Given that water loses 30% of its intensity. Since rate is decreasing so we use b=1-0.3=0.7

And final amount $\dpi{120} A=\frac{1}{2}A_{0}$

$\dpi{120} \frac{1}{2}A_{0}=A_{0}\left ( 0.7 \right )^{\frac{t}{1}}$ here t represents the depth

$\dpi{120} \frac{1}{2}=0.7^{t}$

$\dpi{120} ln\frac{1}{2}=ln0.7^{t}$

ln(0.5) =t ln(0.7)

$\dpi{120} \frac{ln(0.5)}{ln(0.7)}=t$

1.94 m =t

Practice problems:

1. Population of a town doubles every 13 years. If 2330 people were present in year 1990, then find the population in year 2020.

2.A radioactive sample has half life of 2 days. How long will it take for only 1/6^th of the sample to remain?

3.Population of a town halves every 12 years. In how many years will 27% of the population have fled?

4.Light passing through murky water loses 26% of its intensity for every meter of water. At what depth will the intensity be 1/3 of what it is at the surface?

Answers: 1) 11536 people

2) 5.17 days

3) 5.45 years

4) 3.65 m

Applications of exponential functions (Compound Interest)

August 11, 2019/by admin_ct

Formula to calculate compound interest is

$\dpi{150} \mathbf{A= P(1+\frac{i}{n})^{t(n)}}$

Where A= Amount, P= principal sum

i = interest rate in decimals

t= time in years

n =number of times interest is compounded

Have a look at the table showing different values of n for common compounding systems.

Compounding system	Value of n
Annually(once per year)	n=1
Semi-annually(twice in a year)	n=2
Quarterly(4 times in a year)	n=4
Monthly (12 times in a year)	n=12
Daily (365 times in a year)	n =365

Example1: A sum of money $5,000 is invested at 7.2% compounded annually for 4 years. Calculate the amount at the end of 4 years.

Solution: Given that p=5,000 , Since interest is compounded annually so we use n=1

Rate(i) =7.2% =0.072

Time(t)= 4 years

Using formula: $A= P(1+\frac{i}{n})^{t(n)}$

$A= 5000(1+\frac{0.072}{1})^{4(1)}$

$A= 5000\left ( 0.072 \right )^{4}$

A = $6603.12

Example2: A sum of $2,300 is invested at 6% compounded monthly for 7 year. Calculate the interest earned at the end of 7 years.

Solution: Since Interest is compounded monthly so we use n=12 .

Rate(i)= 6%= 0.06

Time(t)= 7 years.

Using formula, $A= P(1+\frac{i}{n})^{t(n)}$

$A= 2300(1+\frac{0.06}{12})^{7(12)}$

$A= 2300(1+0.005)^{84}$

A = 3496.85

Interest earned = Amount – Principal

= 3496.85- 2300= $1196.85

Example3: A sum of $300 is invested at 10% compounded quarterly. How many years must it stay in the bank to double?

Solution: Given principal sum P=$300

And amount(A) = double(600)

Rate(i) = 10% =0.1

Since interest is compounded quarterly so we use n= 4

Using formula, $A= P(1+\frac{i}{n})^{t(n)}$

$600= 300(1+\frac{0.1}{4})^{t(4)}$

$\frac{600}{300}= (1+0.025)^{4t}$

$2= (1.025)^{4t}$

$\ln 2= \ln (1.025)^{4t}$ (taking ln of both sides)

$\ln 2= 4t \ln (1.025)$

$\frac{ln2}{ln(1.025)}= 4t$

28.07 = 4t

7.01 = t

That means money must stay in the bank for 7 years.

Example4: How much money would you need to deposit today at 9% annual interest compounded monthly to have $12000 in the account after 6 years?

Solution: Here we are given amount A=12,000 and we need to find sum invested(p).

Given rate (i) = 9%= 0.09

Time(t)= 6 yrs.

Using formula, $A= P(1+\frac{i}{n})^{t(n)}$

$12000= P(1+\frac{0.09}{12})^{6(12)}$

$12000= P(1+0.0075)^{72}$

12,000 =P(1.712552)

$\frac{12000}{1.712552}= p$

7,007.08 = P

So we need to deposit $7,007.08 today, to get $12000 in 6 years.

Practice problems:

A sum of money $8,000 is invested at 5.6% compounded semi annually for 7 years. Calculate the amount at the end of 7 years.
A sum of $380 is invested at 5% compounded quarterly. How many years must it stay in the bank to triple? Round to nearest year.
How much money would you need to deposit today at 5% annual interest compounded monthly to have $20000 in the account after 9 years?

Answers:

$11775.89
22 years
$12764.49

Solving Exponential Equations

August 10, 2019/by admin_ct

How to solve Exponential Equations

Before learning how to solve exponential function, we should be familiar with basic exponential rules.

$\dpi{120} \mathbf{x^{a}*x^{b}=x^{a+b}}$ powers get added when same bases multiplied
$\dpi{120} \mathbf{\frac{x^{a}}{x^{b}}= x^{a-b}}$ powers get subtracted when same bases divided
$\dpi{120} \mathbf{x^{a}*y^{a}=\left ( xy \right )^{a}}$ having same powers, different bases get multiplied
$\dpi{120} \mathbf{\left ( x^{a} \right )^{b}= x^{ab}}$
$\dpi{120} \mathbf{x^{-k}= \frac{1}{x^{k}}}$ negative powers can be written as fraction

Lets look at one important fact which we are going to use for solving exponential functions,

$\dpi{120} \mathbf{a^{x}=a^{y} \Rightarrow x=y}$

When bases are same, we can equate the exponents(powers).

Lets look at some easy problems first and understand the process with the help of video lesson.

1 Example: Solve the given exponential equation for x.

$\dpi{120} {\color{Red} 2^{2x-1}=2^{x+3}}$

Solution: As we can see, both sides of this equation have same bases, so we can equate the powers and solve for x.

2x-1 = x+3

2x-x = 3+1

x = 4

2Example: Solve the following exponential function for x.

$\dpi{120} {\color{Red} 4^{3x}= 8^{x+1}}$

Solution: Since bases are not same so we have to make them same. We know that both 4 and 8 can be written using exponents of 2.

$\dpi{120} \left ( 2^{2} \right )^{3x}=\left ( 2^{3} \right )^{x+1}$

$\dpi{120} 2^{6x}= 2^{3x+3}$ distribute , 3(x+1)=3x+3

6x = 3x+3 For same bases, power can be equated

3x = 3

x = 1

3 Example: solve for x , $\dpi{120} {\color{Red} 4^{5-9x}= \frac{1}{8^{x-2}}}$

Solution: $\dpi{120} 4^{5-9x}= \frac{1}{8^{x-2}}$

$\dpi{120} 4^{5-9x}= 8^{-(x-2)}$ rewrite fraction using negative exponents

$\dpi{120} \left ( 2^{2} \right )^{5-9x}= \left ( 2^{3} \right )^{-(x-2)}$

$\dpi{120} 2^{2(5-9x)}=2^{3(-x+2)}$ bases made similar

$\dpi{120} 2^{10-18x}=2^{-3x+6}$

10 -18x = -3x +6 having same bases, powers get equated

10-6 = -3x+18x

4 = 15x => x = 4/15

4.Example : Solve the given exponential for x.

$\dpi{120} {\color{Red} 3^{x^{2}}= 3^{6-x}}$

Solution: Nothing much to do as we have already same bases.

So we just equate the exponents and solve for x.

$\dpi{120} x^{2}= 6-x$

$\dpi{120} x^{2}+x-6=0$

(x+3)(x-2) =0

x+3 =0 , x-2 =0

x= -3 , x=2

So far we have worked on examples where we were able to get same bases on both sides of the equation, but it is not always the case. In some exponential equations, we are not given same bases and we are not able to make them same anymore for example $\dpi{120} 5^{x+1}=3^{2}$ . In such cases we work on them taking natural log (ln) of both sides. Lets work on such examples and understand the process with video lesson.

5.Example: Solve the following exponential equations for x. Round your answers to nearest thousandth.

$\dpi{120} {\color{Red} 5^{x+1}=3^{2}}$
$\dpi{120} {\color{Red} 2^{x+1}= 5^{1-2x}}$
$\dpi{120} {\color{Red} 0.3^{1+x}= 1.7^{2x-1}}$
$\dpi{120} {\color{Red} e^{x-5}= 100}$

Solution:1. $\dpi{120} 5^{x+1}=3^{2}$

Taking ln of both sides,

$\dpi{120} \ln 5^{x+1}= \ln 3^{2}$

(x+1)ln5 = ln9 apply power formula of logs

$\dpi{120} x+1 = \frac{ln9}{ln5}$ Please note that $\dpi{100} \frac{ln9}{ln5}\neq ln \left ( \frac{9}{5} \right )$

x+1 = 1.365

x = 0.365

2. $\dpi{120} 2^{x+1}= 5^{1-2x}$

Taking ln of both sides,

$\dpi{120} \ln 2^{x+1}= \ln 5^{1-2x}$

(x+1)ln2 = (1-2x)ln5

xln2 +ln2 = ln5-2xln5

xln2+2xln5 = ln5 –ln2

x(ln2+2ln5)= ln(5/2)

x(ln2+ln5^2 ) = ln(2.5)

x(ln2*25) = ln(2.5)

$\dpi{120} x= \frac{ln(2.5)}{ln(50)} = 0.234$

3. $\dpi{120} 0.3^{1+x}= 1.7^{2x-1}$

Taking ln of both sides,

$\dpi{120} \ln 0.3^{1+x}= \ln 1.7^{2x-1}$

(1+x)ln(0.3) =(2x-1) ln(1.7) Used power rule of log

ln(0.3)+xln(0.3) = 2xln(1.7) –ln(1.7)

ln(0.3) +ln(1.7) = 2xln(1.7)-xln(0.3)

ln(0.3 *1.7) = x[2ln(1.7)-ln(0.3)]

$\dpi{120} \frac{ln(0.51)}{2ln(1.7)-ln(0.3)} = x$

$\dpi{120} \frac{ln(0.51)}{ln(1.7)^{2}-ln(0.3)} = x$

$\dpi{120} \frac{ln(0.51)}{ln\left ( \frac{1.7^{2}}{0.3} \right )}=x$

$\dpi{120} \frac{ln(0.51)}{ln(9.633)}=x$

4. $\dpi{120} e^{x-5}=100$

taking ln of both sides,

$\dpi{120} e^{x-5}=100$

(x-5)ln(e) = ln(100) Since ln and e are inverse of each other so ln(e)= 1

(x-5) = ln(100)

x = 5+ ln(100)

x = 9.605

Practice problems:

Solve the following exponential equations. Round the answer to nearest thousandth wherever possible.

1. $\dpi{120} 3^{2x+1}=81$

2. $\dpi{120} 2^{2x+2}-1 = 15$

3. $\dpi{120} e^{3x}=30$

4. $\dpi{120} \left ( \frac{4}{3} \right )^{1-x}=5^{x}$

Answers:

x=3/2
x=1
x=1.134
x=0.152

Simplify Exponential expressions

August 9, 2019/by admin_ct

How to simplify exponential expressions.

To learn the simplification of expressions with exponents, we should be aware about the rules of exponents first.

$\dpi{120} \mathbf{Product Rule} :\mathbf{x^{a}*x^{b}=x^{a+b}}$ powers get added when same bases multiplied
$\dpi{120} \mathbf{Quotient Rule }: \mathbf{\frac{x^{a}}{x^{b}}=x^{a-b}}$ powers get subtracted when same bases divided
$\dpi{120} \mathbf{Power Rule}: \mathbf{\left ( x^{a} \right )^{b}= x^{ab}}$
$\dpi{120} \mathbf{\mathbf{\mathbf{}}Common Power Rule: \left ( x^{a} \right )\left ( y^{a} \right )= \left ( xy \right )^{a}}$ having same powers, different bases get multiplied
$\dpi{120} \mathbf{Negative Exponent Rule}: \mathbf{x^{-1}=\frac{1}{x}}$ negative powers can be written as fractions
$\dpi{120} \mathbf{x^{-k}=\frac{1}{x^{k}}}$
$\dpi{120} \mathbf{x^{0}=1}$

There may be many ways to simplify an expression using rules of exponents and still ending up with same result (answer). Let’s understand simplification of exponential expressions using following examples.

1. Simplify $\dpi{120} {\color{Red} \left ( 4x^{2} \right )\left ( 3x^{3} \right )}$

$\dpi{120} 4*3*x^{2}*x^{3}$

$\dpi{120} 12x^{2+3} = 12x^{5}$

2. Simplify $\dpi{120} {\color{Red} \left (\frac{24x^{5}}{3x^{2}} \right )}$

$\dpi{120} \left ( \frac{24}{3} \right )\left ( \frac{x^{5}}{x^{2}} \right )= \left (8 \right )(x^{5-2})=8x^{3}$ (write numbers and variables terms together and divide using exponent’s rules)

3. Simplify $\dpi{120} {\color{Red} \left ( 5x^{3} \right )^{4}}$

$\dpi{120} \left ( 5x^{3} \right )^{4}= 5^{4}\left ( x^{3} \right )^{4}= 625x^{12}$ (power get distributed to numerical and variable parts

4. Simplify $\dpi{120} {\color{Red} \left ( x^{-2}y^{2} \right )^{5}\left (x^{5} y \right )^2}$

$\dpi{120} \left ( x^{-2} \right )^{5}\left ( y^{2} \right )^{5}\left ( x^{5} \right )^{2}y^{2}$ (distributing powers to each element)

$\dpi{120} \left ( x^{-10} \right )\left ( y^{10} \right )\left ( x^{10} \right )\left ( y^{2} \right )$

$\dpi{120} \left ( x^{-10+10} \right )\left ( y^{10+2} \right )$ (exponents of same bases combined)

$\dpi{120} x^{0}y^{12}=1 y^{12}=y^{12}$

5. Simplify $\dpi{120} {\color{Red} \left ( \frac{x^{2}y}{z} \right )\left ( \frac{x^{3}z}{y^{3}} \right )}$

$\dpi{120} \left ( \frac{x^{2}y}{z} \right )\left ( \frac{x^{3}z}{y^{3}} \right )$ = $\dpi{120} \left ( x^{2}y \right )\left ( \frac{x^{3}}{y^{3}} \right )$ (cross cancelling z)

$\dpi{120} \left ( x^{2} \right )\left ( x^{3} \right )\left ( \frac{y}{y^{3}} \right )$ ( writing like terms together)

$\dpi{120} \left ( x^{2+3} \right )\left ( y^{1-3} \right )$ (applying rules of exponents)

$\dpi{120} \left ( x^{5} \right )\left ( y^{-2} \right )= \frac{x^{5}}{y^{2}}$

In the following video, we are going to learn how to simplify complex expressions with exponents.

Practice Problems:

Simplify using rules of exponents:

1. $\dpi{120} \left ( x^{2}y \right )^{3}\left ( xy^{2} \right )^{2}$

2. $\dpi{120} \left ( 4x \right )^{3}\left ( 2x \right )^{-5}$

3. $\dpi{120} \frac{2x^{4}y}{\left ( 3xy \right )^{2}}$

4. $\dpi{120} \left ( \frac{x^{2}}{y} \right )^{3}\left ( \frac{y}{x} \right )^{2}$

Answers:

1. $\dpi{120} x^{8}y^{7}$

2. $\dpi{120} \frac{2}{x^{2}}$

3. $\dpi{120} \frac{2x^{2}}{9y}$

4. $\dpi{120} \frac{x^{4}}{y}$

Basics of Exponential and Logarithmic functions

August 9, 2019/by admin_ct

Exponential function:

Any function represented as $\dpi{120} f(x)= a^{x}$ is called exponential function, where a>0 and a≠1 .

Domain of an exponential function is R, set of all real numbers and range is (0,∞ ) because exponential function attains only positive values.

As a>0 and a≠1 so we have the following cases.

Case 1. When a >1

Then exponential function is always an increasing function with y intercept as (0,1).

Case 2. When 0<a<1

Then exponential function is always a decreasing function with same y intercept as(0,1).

Following graphs will help you to observe the similarity and difference between two cases.

Case 3: When a=1

Then $\dpi{120} f(x)=1^{x}=1$ which is a constant function.

Whatever be the value of a, y intercept is always (0,1) is every case.

Logarithmic function:

The function represented as $\dpi{120} f(x)=\log_{a}x$ , a>0, and a ≠1 for x>0 is called logarithmic function.

Log functions are never defined for 0 and negative values. Domain of logarithmic functions is set of all non negative real numbers i.e. (0,∞ ). Range is the set of all real numbers.

As logarithmic and exponential functions are inverse of each other so their domain and range are interchanged.

Depending on the values of a, following are the possible cases.

Case 1. When a>1

Then logarithmic function is increasing function with x intercept as (1,0).

Case 2. When 0<a<1

Then logarithmic function is decreasing function with x intercept same as (1,0).

When base of log is constant e(2.718) instead of a , $\dpi{120} f(x)=\log_{e}x$ then we call the logarithmic function as natural log function f(x)=ln(x).

Following graph will help you to observe the similarity and difference between two cases.

Relationship between logarithmic and exponential function.

We can observe the relationship between exponential and logarithmic function by looking at their graphs together.

Clearly logarithmic and exponential functions are inverse functions!

Logarithmic function is the reflection of exponential function in the line y=x. In other words , axes have been swapped , x become f(x) and f(x) becomes x.

Simplify Radical expressions

August 8, 2019/by admin_ct

Simplify algebraic radical expressions

To learn simplification of radical expressions, we first need to know some basic rules. We can write radicals using exponents as follows:

$\dpi{120} \sqrt{x}= x^{\frac{1}{2}}$

$\dpi{120} \sqrt[3]{x}= x^{\frac{1}{3}}$

$\dpi{120} \sqrt[4]{x}= x^{\frac{1}{4}}$ and so on…..

So basically we can write $\dpi{120} \sqrt[a]{x}= x^{\frac{1}{a}}$

Some more rules :

1. $\dpi{120} \sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}$ or $\dpi{120} \left ( \frac{x}{y} \right )^{a}=\frac{x^{a}}{y^{a}}$

2. $\dpi{120} \sqrt{xy}=\sqrt{x}\sqrt{y}$ or $\dpi{120} \left ( xy \right )^{a}=x^{a}y^{a}$

To simplify nth root radicals, first we write the numerical part in form of their prime factors and then we divide the exponents by nth root (index). Whole numbers go outside radical and remainder remain inside radical. To understand this process lets see few examples.

Example1. Simplify $\dpi{120} {\color{Red} \sqrt[3]{24x^{3}y^{4}}}$

Solution:

Step (i) write numerical part (24) in form of prime factors. And cube root as exponent 1/3

$\dpi{120} \left ( 2^{3}*3*x^{3}*y^{3} \right )^{\frac{1}{3}}$

Step (ii) divide the exponents by 3 write whole numbers outside brackets and remainders inside.

$\dpi{120} 2xy\left ( \3y \right )^{\frac{1}{3}}$ Final answer

Example2. Simplify $\dpi{120} {\color{Red} \sqrt[4]{16x^{8}y^{4}}}$

Step (i) Write nth root in exponent form and numerical part as product of exponential prime factors

$\dpi{120} \left ( 2^{4}x^{8}y^{4} \right )^{\frac{1}{4}}$

Step (ii) Divide all the exponent by 4 . Here remainder is 0 for each one so fourth root goes off.

$\dpi{120} 2x^{2}y$ Final answer

Example3. Simplify $\dpi{120} {\color{Red} \sqrt{\frac{150}{21}}}$

Solution:

$\dpi{120} \sqrt{\frac{150}{21}} = \frac{\sqrt{2*3*5*5}}{\sqrt{3*7}}$

$\dpi{120} \frac{\sqrt{2*3*5^{2}}}{\sqrt{3*7}}$

$\dpi{120} \frac{5*\sqrt{2}*\sqrt{3}}{\sqrt{3}*\sqrt{7}}$

$\dpi{120} \frac{5\sqrt{2}}{\sqrt{7}}$

Other way: Reduce 150/21 to lower term by dividing by 3

$\dpi{120} \sqrt{\frac{150}{21} }= \sqrt{\frac{50}{7}}= \frac{\sqrt{2*5*5}}{\sqrt{7}}= \frac{5\sqrt{2}}{\sqrt{7}}$

Example4. Simplify $\dpi{120} {\color{Red} \sqrt[3]{\frac{27}{8}}}$

Solution: $\dpi{120} \left ( \frac{27}{8} \right )^{\frac{1}{3}}=\frac{27^{\frac{1}{3}}}{8^{\frac{1}{3}}}$

$\dpi{120} \frac{\left ( 3^{3} \right )^{\frac{1}{3}}}{\left ( 2^{3} \right )^{\frac{1}{3}}}= \frac{3}{2}$

Practice problems:

Simplify :

$\dpi{120} \frac{\sqrt{32}}{4}$

2. $\dpi{120} \frac{\sqrt{12}}{\sqrt{3}}$

3. $\dpi{120} \sqrt[5]{32x^{5}y^{10}}$

4. $\dpi{120} \sqrt{18x^{4}y^{6}}$

5. √18 *√2

Answers:

√2
2
$\dpi{120} 2xy^{2}$
$\dpi{120} 3\sqrt{2}x^{2}y^{3}$
6

Operations on radical expressions

August 8, 2019/by admin_ct

Addition , Subtraction of radical expressions

Only like radical can be added or subtracted. So first we should know what are like radicals.

Like Radicals:

These are the expressions which have same radical part (radicand) or they can be simplified further to get same radical parts.

For example √2, 3√2, 10√2 and -5√2 are all like radicals.

On the other hand √32 and √18 are not like radicals but they can be simplified further to get like radicals.

Lets see how √32 and √18 can be simplified.

$\dpi{120} \sqrt{32}= \sqrt{2*4*4}=\sqrt{2*4^{2}}= 4\sqrt{2}$

$\dpi{120} \sqrt{18}= \sqrt{2*3*3}=\sqrt{2*3^{2}}= 3\sqrt{2}$

Now both are like radicals because both have same radical part (√2 ) and they can be added now.

Only the coefficients of like radicals are combined, keeping radical parts as it is.

$\dpi{120} \sqrt{32}+\sqrt{18}= 4\sqrt{2}+3\sqrt{2} = (4+3)\sqrt{2}= 7\sqrt{2}$

Example1. Combine the given radical expressions.

$\dpi{120} {\color{Red} 3\sqrt{5}-\sqrt{2}+\sqrt{5}+5\sqrt{2}}$

Solution: $\dpi{120} 3\sqrt{5}+\sqrt{5}+5\sqrt{2}-\sqrt{2}$ ( write like radical terms together)

$\dpi{120} (3+1)\sqrt{5}+(5-1)\sqrt{2}$ (combined coefficient of like radicals

$\dpi{120} 4\sqrt{5}+4\sqrt{2}$

Some radicals get simplified completely !

Example2. Combine the following radical expression

$\dpi{120} {\color{Red} \sqrt{121}-\sqrt{64}}$

Solution: $\dpi{120} \sqrt{11*11}-\sqrt{8*8}$

$\dpi{120} \sqrt{11^{2}}-\sqrt{8^{2}}$

11 – 8 = 3

Before combining the radical expressions , ensure all radicals are simplified !

Example3. Combine the following radical expression

$\dpi{120} {\color{Red} 5\sqrt{7}-2\sqrt{28}+\sqrt{7}}$

Solution: $\dpi{120} 5\sqrt{7}-2\sqrt{2*2*7}+\sqrt{7}$

$\dpi{120} 5\sqrt{7}-2\sqrt{2^{2}*7}+\sqrt{7}$

$\dpi{120} 5\sqrt{7}-2*2\sqrt{7}+\sqrt{7}$

$\dpi{120} 5\sqrt{7}-4\sqrt{7}+1\sqrt{7}$ [ In absence of any coefficient, 1 can be used just like x can be written as 1x]

$\dpi{120} (5-4+1)\sqrt{7}= 2\sqrt{7}$

But there is no such restriction while multiplying or dividing radicals. We can multiply or divide any radicals together.

For example

Unlike radicals : Like radicals :

√2*√3=√6 √2*√2=2

√5*√7=√35 √7*√7= 7

$\frac{\sqrt{21}}{\sqrt{7}}= \frac{\sqrt{7}*\sqrt{3}}{\sqrt{7}} =\sqrt{3}$ $\frac{2\sqrt{3}}{\sqrt{3}}= 2$

$\frac{\sqrt{45}}{\sqrt{5}}= \frac{\sqrt{9}*\sqrt{5}}{\sqrt{5}}=\sqrt{9}=3$ $\frac{\sqrt{7}}{\sqrt{7}}=1$

Example4. Simplify and combine the following radical expressions

${\color{Red} \sqrt{3}*\sqrt{2}-2\left ( \sqrt{18}\div \sqrt{3} \right )+3\sqrt{6}}$

So0lution:

$\sqrt{3}*\sqrt{2}-2\left ( \sqrt{18}\div \sqrt{3} \right )+3\sqrt{6}$

$\sqrt{6}-2(\sqrt{6})+3\sqrt{6}$

$1\sqrt{6}-2\sqrt{6}+3\sqrt{6}$

$(1-2+3)\sqrt{6}= 2\sqrt{6}$

Practice problems:

1. $\dpi{120} 5\sqrt{3}+\sqrt{3}+5\sqrt{3}-\sqrt{3}$

2. $\dpi{120} \sqrt{75}+\sqrt{48}$

3. $3\sqrt{6}+\sqrt{3}*\sqrt{2}$

4. $5\sqrt{3}*\sqrt{7}-2\sqrt{21}$

Answers:

10√3
9√3
4√6
3√21

Solving Linear equation

August 7, 2019/by admin_ct

How to solve linear equation in one variable

The word equation is derived from word ‘equal’. Any expression having equal (=) sign is called equation. If any linear algebraic expression in one variable is equal to some integer or some other algebraic expression, then it is called algebraic equation.
Linear equations can be categorized as follows
1. When variable is only one side.
2. When variable is on both sides

When variable is only on one side:

We always solve the equation for an unknown variable using the same operations on both sides so that equation doesn’t become uneven and both sides remain balanced. Lets look at some examples of multi step equations starting from single step equations.

Example1. Solve $\dpi{120} {\color{Red} x+\frac{4}{5}=\frac{1}{2}}$

Solution: Our ultimate aim is to isolate x here. So we need to get rid of 4/5 , for that we subtract 4/5 from both sides. $\dpi{120} x+\frac{4}{5}=\frac{1}{2}$

$\dpi{120} x+\frac{4}{5}-\frac{4}{5} = \frac{1}{2}-\frac{4}{5}$ [ subtracting 4/5 from both sides]

$\dpi{120} x = \frac{1}{2}-\frac{4}{5}$

$\dpi{120} x = \frac{5}{10}-\frac{8}{10}= \frac{-3}{10}$

Example 2. Solve $\dpi{120} {\color{Red} \frac{x}{3}-5 =-2}$

Solution: We move close to x step by step. For that first we add 5 to both sides.

$\dpi{120} \frac{x}{3}-5+5=-2+5$

$\dpi{120} \frac{x}{3}=3$

Using the reverse operation, multiply both sides with 3

$\dpi{120} \frac{x}{3}(3)=3(3)$

x = 9

Example3. Solve 3(x-1) +2 = 14
Solution: This equation can be solved using two ways. Either you can first distribute 3 and combine like terms on left before isolating x, or you can first subtract 2 and then divide both sides by 3. But if you get fraction on dividing then better to skip this way and use other way instead.

3(x-1) +2 = 14
3(x-1) = 12 [subtracted 2 from both sides]

$\frac{3(x-1)}{3}= \frac{12}{3}$

x -1 = 4

x = 5

When variable is on both sides:

When variables on both sides of equation, we try to bring the variables together either on left or on right side and then move forward with the same process of isolating variable.

Example4. Solve for x, 3x-5 = x+3

Solution: 3x-5 =x+3

3x-x -5 = x-x+3 [ subtracting x from both sides]

2x-5 = 3

2x-5+5 = 3+5 [ adding 5 on both sides]

2x = 8

x = 8/2 = 4

Example5. Solve ${\color{Red} \frac{3x}{2}-\frac{7}{2}=\frac{5x}{2}+\frac{1}{4}}$

Solution: $\frac{3x}{2}-\frac{5x}{2}-\frac{7}{2}=\frac{5x}{2}-\frac{5x}{2}+\frac{1}{4}$ [subtracted 5x/2 from both sides]\

$\frac{-2x}{2}-\frac{7}{2}=\frac{1}{4}$

$-x = \frac{1}{4}+\frac{7}{2}$

$-x = \frac{1}{4}+\frac{14}{4} = \frac{15}{4}$

x = -15/4

Example6. Solve : 5-3(x-1) = 2(3x-5)
Solution: Distribute -3 on left and 2 on right, we get,
5- 3(x)-3(-1) = 2(3x) +2(-5)
5-3x+3 = 6x -10
-3x + 8 = 6x-10
-3x-6x = -10-8
– 9x = -18

$\frac{-9x}{-9}=\frac{-18}{-9}$

x = 2

Word problems on linear equations in one variable:

Most students get overwhelmed by word problems but these are not hard if each word is interpreted correctly. Lets work on some examples .

Example6.Twice of a number when increased by 3 becomes 9. Find the
number.
Solution: First we assume unknown number as x. Now we get the words like this.
Twice of x , increased by 3 becomes 9.
2x + 3 = 9
Replace words ‘increased by’ with + and ‘becomes’ with = sign.
We got the equation and now solve it just like we solve any linear
equation.
2x + 3 = 9
2x+3-3=9-3 [subtracting 3 from both sides]
2x = 6
x = 6/2 = 3

Example7. In a martial arts class, number of girls were ten more than twice the number of boys. If there were total 85 students, then find number of boys and girls in the class.
Solution: First we assume the unknown variable as x. Since information about girls are dependent on boys so we assume number of boys as x.
Let number of boys = x
Number of girls = ten more than twice the number of boys
Number of girls = 10 + twice (x)
Number of girls = 10 + 2x
boys + girls = total students
x + 10+2x = 85
3x+10 = 85
3x = 85-10
3? = 75
x =75/3 = 25

So boys = 25

girls = 10+2(25)= 60

Example8. After an increment of 25% , Maria got her monthly salary as $4000. Find her salary per month before increment.
Solution: We generally assume what we need to find. So lets assume Maria’s salary before increment = x
Original salary + increment = new salary
x + 25% of x = 4000
x + 0.25x = 4000
1.25 x = 4000
1.25?/1.25 = 4000/1.25 = 3200
x= 3200
Maria’s salary before increment was $3200

Practice problems:

Solve the following linear equations for x.
1. 7x – 2 = 3/2
2. 5+3(x+2) = 20
3. 9-(x+3) = 5(x-1)+3
4. Sum of two numbers is 25. If one number is one fourth of other ,
then find both the numbers
5. Mark bought some eatables to celebrate his birthday. He bought
Twice as many hot dogs as pizzas and one third as many burgers
as hot dogs. If he bought total 22 eatable items then find each
kind of food he bought.

Answers:
1. 1/2
2. 3
3. 4/3
4. 20, 5
5. Pizzas= 6 ,Hot dogs =12 ,Burgers = 4

Sum of n terms of Geometric Sequence

August 6, 2019/by admin_ct

Sum of n terms of Geometric Sequence

Sum of n terms of a finite geometric sequence is given by two formulas depending on the values of common ratio(r).

Let ‘a’ be the first term of geometric sequence and ‘r’ be its common ratio then sum of n terms, is given as,

$\dpi{150} \mathbf{S_{n}= a\left ( \frac{r^{n}-1}{r-1} \right )} for \left | r \right |>1$

$\dpi{150} \mathbf{S_{n}= a\left ( \frac{1-r^{n}}{1-r} \right )} for \left | r \right |< 1$

Example1. Consider the geometric series -90+30+(-10)+10/3 +….

Find the sum of first 16 terms .
Find n such that $\dpi{120} {\color{Red} S_{n}= -66.67}$

Solution: a) First term(a) = -90

Common ratio(r) = -1/3

Here |r| < 1 so we use second formula.

Sum of first 16 terms $\dpi{120} S_{16}= -90\left ( \frac{1-\left ( \frac{-1}{3} \right )^{16}}{1-\left ( \frac{-1}{3} \right )} \right )$

$\dpi{120} S_{16}= -90\left ( \frac{1-\frac{1}{3^{16}}}{1+\frac{1}{3}} \right )$

$\dpi{120} S_{16}= -90\left ( \frac{\frac{3^{16}-1}{3^{16}}}{\frac{4}{3}} \right )$

$\dpi{120} S_{16}= -90\left ( \frac{3^{16}-1}{3^{16}}*\frac{3}{4} \right )$

$\dpi{120} S_{16}= -90\left ( \frac{43046720}{3^{15}*4} \right )$

$\dpi{120} S_{16}= \left ( \frac{-107616800}{1594323} \right ) \Rightarrow 67.4999 \approx 67.5$

b) Given that $\dpi{120} S_{n}= -66.67$

$\dpi{120} S_{n}= -\frac{200}{3}$

$\dpi{120} -90\left ( \frac{1-\left ( \frac{-1}{3} \right )^{n}}{1-\left ( \frac{-1}{3} \right )} \right ) = \frac{-200}{3}$

$\dpi{120} \left ( \frac{1-(\frac{-1}{3})^{n}}{1+\frac{1}{3}} \right )= \frac{-200}{-90*3}$

$\dpi{120} \left ( \frac{1-(\frac{-1}{3})^{n}}{\frac{4}{3}} \right )= \frac{20}{27}$

$\dpi{120} 1-\left ( \frac{-1}{3} \right )^{n} = \frac{20}{27}*\frac{4}{3}= \frac{80}{81}$

$\dpi{120} 1-\frac{80}{81}= \left ( \frac{-1}{3} \right )^{n}$

$\dpi{120} \frac{1}{81}=\left ( \frac{-1}{3} \right )^{n}$

$\dpi{120} \left ( \frac{-1}{3} \right )^{4}=\left ( \frac{-1}{3} \right )^{n}$ [ For same bases , exponents can be equated]

4 = n

Example2: Find sum of given finite geometric series.

$\dpi{120} {\color{Red} \sum_{n=1}^{10}8\left ( \frac{3}{4} \right )^{n-1}}$

Solution: Here we are going to find $\dpi{120} s_{10}$ for the given series

$\dpi{120} 8\left [ 1+\frac{3}{4}+\left ( \frac{3}{4} \right )^{2}+............+\left ( \frac{3}{4} \right )^{9} \right ]$

$\dpi{120} 8+6+\frac{9}{2} +......+ \left ( \frac{3}{4} \right )^{9}$

First term(a) = 8

Common ratio(r) = 3/4

$\dpi{120} S_{10}= 8\left ( \frac{1-\left ( \frac{3}{4} \right )^{10}}{1-\left ( \frac{3}{4} \right )} \right )$

$\dpi{120} = 8\left ( \frac{0.943686}{0.25} \right )=30.20$

Example3: Find the sum of following series.

5+55+555+….n terms

Solution: Let $\dpi{120} S_{n}$ be the sum of given series then,

$\dpi{120} S_{n}$ = 5+55+555+……..n terms

= 5[1+11+111+…..n terms]

= 5/9[9+99+999+……. n terms]

= 5/9 [(10-1)+(100-1)+(1000-1) +…….n terms]

= $\dpi{120} \frac{5}{9}\left [ (10-1)+(10^{2}-1)+\left ( 10^{3}-1 \right )+.....n terms \right ]$

= $\dpi{120} \frac{5}{9}\left [\left (10+10^{2}+10^{3}+.....+10^{n} \right )-(1+1+1+....n times) \right ]$

= $\dpi{120} \frac{5}{9}\left [ 10\left ( \frac{10^{n}-1}{10-1} \right )-n \right ]$

= $\dpi{120} \frac{5}{9}\left [ 10 \frac{10^{n}-1}{9} -n \right ]$

= $\dpi{120} \frac{5}{81}\left [ 10(10^{n}-1) -9n\right ]$

= $\dpi{120} \frac{5}{81}\left [ 10^{n+1}-10-9n \right ]$

Example4: Sum of first three numbers in a G.P. is 16 and sum of next three terms is 128. Find the sum of n terms .

Solution: Let a be the first term and r be the common ratio of G.P. Then,

$\dpi{120} a+ar+ar^{2}=16$

$\dpi{120} a\left (1+r+r^{2} \right )=16$ ———(i)

Also $\dpi{120} ar^{3}+ar^{4}+ar^{5}=128$

$\dpi{120} ar^{3}\left (1+r+r^{2} \right )=128$ ——–(ii)

Dividing equation (ii) by equation(i) we get,

$\dpi{120} \frac{ar^{3}\left ( 1+r+r^{2} \right )}{a(1+r+r^{2})} = \frac{128}{16}$

$\dpi{120} \frac{ar^{3}}{a}= 8$

$\dpi{120} r^{3}=8$

r = 2

Substitute r=2 into equation (i) we get,

$\dpi{120} a\left (1+2+2^{2} \right )=16$

a(7) = 16

a = 16/7

And sum is given as, $\dpi{120} S_{n}= a\left ( \frac{r^{n}-1}{r-1} \right )$

$\dpi{120} S_{n}= \frac{16}{7}\left ( \frac{2^{n}-1}{2-1} \right )= \frac{16}{7}\left ( 2^{n}-1 \right )$

Example 5: A G.P. consists of even number of terms. If the sum of all terms is 5 times the sum of the terms occupying the odd places. Find the common ratio r.

Solution: Let there be 2n(even) terms in G.P with first term as a and common ratio as r.

Given that,

Sum of all terms = 5(sum of terms occupying the odd places)

$\dpi{120} a_{1}+a_{2}+a_{3}+.....+a_{2n}=5\left ( a_{1}+a_{3}+a_{5}+a_{2n-1} \right )$

$\dpi{120} a+ar+ar^{2}+....+ar^{2n-1}=5\left ( a+ar^{2}+ar^{4}+....+ar^{2n-2} \right )$

$\dpi{120} a\left ( \frac{1-r^{2n}}{1-r} \right )=5a\left ( \frac{1-\left ( r^{2} \right )^{n}}{1-r^{2}} \right )$

Note: if there are total 2n terms then there would be n even terms and n odd terms.

$\dpi{120} a\left ( \frac{1-r^{2n}}{1-r} \right )=5a\left ( \frac{1-r^{2n}}{\left (1-r \right )\left ( 1+r \right )} \right )$

Canceling $\dpi{120} \left ( 1-r^{2n} \right )$ from numerators and (1-r) from denominators on both sides, we get

$\dpi{120} a=\frac{5a}{1+r}$

a(1+r) = 5a

1+r = 5

r = 4

Sum of Infinite G.P.

Sum of infinite G.P with a as first term and r as common ratio is given as,

$\dpi{150} \mathbf{S_{\infty }= \frac{a}{1-r}}$

Example6: Find the sum of following geometric infinite series

${\color{Red} \frac{1}{2}+\frac{1}{3^{2}}+\frac{1}{2^{3}}+\frac{1}{3^{4}}+\frac{1}{2^{5}}+......\infty }$

Solution: We can separate these terms to get two infinite geometric series.

$\left ( \frac{1}{2}+\frac{1}{2^{3}}+\frac{1}{2^{5}}+...... \right )+\left ( \frac{1}{3^{2}}+\frac{1}{3^{4}}+\frac{1}{3^{6}}...... \right )$

(Infinite G.P with a=1/2, r=1/4) + (infinite G.P with a=1/9 , r=1/9)

$\dpi{120} \left ( \frac{\frac{1}{2}}{1-\frac{1}{2^{2}}} \right )+\left ( \frac{\frac{1}{3^{2}}}{1-\frac{1}{3^{2}}} \right )$

$\dpi{120} \left ( \frac{\frac{1}{2}}{1-\frac{1}{4}} \right )+\left ( \frac{\frac{1}{9}}{1-\frac{1}{9}} \right )$

$\dpi{120} \frac{1}{2}*\frac{4}{3}+\frac{1}{9}*\frac{9}{8}= \frac{2}{3}+\frac{1}{8}= \frac{19}{24}$

Example 7: prove that $\dpi{120} {\color{Red} 6^{\frac{1}{2}}*6^{\frac{1}{4}}*6^{\frac{1}{8}}......\infty =6}$

Solution: $\dpi{120} 6^{\frac{1}{2}}*6^{\frac{1}{4}}*6^{\frac{1}{8}}......\infty = 6^{\left [ \frac{1}{2}+\frac{1}{4}+\frac{1}{8}.....\infty \right ]}$

Here $\dpi{120} \left [ \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....\infty \right ]$ is a infinite geometric series so we can use formula $S_{\infty }= \frac{a}{1-r}$

$\dpi{120} 6^{\frac{1}{2}}*6^{\frac{1}{4}}*6^{\frac{1}{8}}......\infty = 6^{\left [ \frac{\frac{1}{2}}{1-\frac{1}{2}} \right ]}$

$= 6^{\left [ \frac{\frac{1}{2}}{\frac{1}{2}} \right ]}=6^{1}=6$

Example8: In 1990 the average monthly bill for cellular telephone service in the United States was $80.90. From 1990 through 1997, the average monthly bill decreased by about 8.6% per year.

On average, what did a person pay for cellular telephone service during 1990–1997?

Solution: Here we have,

Common ratio(r) = 1-0.086=0.914

Average monthly bill =$80.90

Average annual bill = 12*80.90=970.8

So first term(a) =970.8

Average cost for cellular telephone service during 8 years is given as, $\dpi{120} S_{n}= a\left ( \frac{1-r^{n}}{1-r} \right )$

$\dpi{120} S_{8}= 970.8\left ( \frac{1-0.914^{8}}{1-0.914} \right )= 5790$

On average a person pay 5790 dollars for cellular telephone service.

Practice problems:

Find the sum of the first 8 terms of the geometric series 1 + 8 + 64 + 512 + . . .
Find sum of following series. $\sum_{n=1}^{6}4 \left ( \frac{3}{2} \right )^{n}$
Find the sum of following series 0.7+0.77+0.777+….n terms
How many terms of the geometric series 3, 3/2 , 3/4 ….be taken together to make 3069/512
Find a G.P. for which sum of first two terms is -4 and the fifth term is 4 times the third term.
A person writes a letter to four of his friends. He asks each of them to copy the letter and mail it to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 cents to mail one letter. Find the amount spent on the postage when 8^th set of letter is mailed.

Answers:

1) 2396745

2) 124.6875

3) $\frac{7}{8}\left [ 9n-1 +\frac{1}{10^{n}} \right ]$

4) 10

5) -4/3 ,-8/3 , -16/3…… or 4,-8,16….

6) $43690

Archives

Intensity of earthquakes and sound loudness

Practice problems:

How to solve Exponential Equations

How to simplify exponential expressions.

Exponential function:

Logarithmic function:

Simplify algebraic radical expressions

Addition , Subtraction of radical expressions

Like Radicals:

How to solve linear equation in one variable

When variable is only on one side:

When variable is on both sides:

Word problems on linear equations in one variable:

Practice problems:

Sum of n terms of Geometric Sequence

Sum of Infinite G.P.