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College Algebra : Exponential and Logarithmic Functions

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Example Question #71 : Exponential And Logarithmic Functions

Solve for x:

$log_{2}x=4$

Possible Answers:

x=16

x=2

x=32

x=8

Correct answer:

x=16

Explanation:

In order to solve for x in a logarithmic function, we need to change it into exponential form. That looks as follows:

$log_{b}x=y\rightarrow b^{y}=x$

For this problem, manipulate the log and solve:

$log_{2}x=4$

$2^{4}=x$

x=16

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Example Question #72 : Exponential And Logarithmic Functions

Solve for x:

$log_{4}x=2$

Possible Answers:

x=8

x=64

x=16

x=2

Correct answer:

x=16

Explanation:

In order to solve for x in a logarithmic function, we need to change it into exponential form. That looks as follows:

$log_{b}x=y\rightarrow b^{y}=x$

For this problem, manipulate the log and solve:

$log_{4}x=2$

$4^{2}=x$

x=16

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Example Question #73 : Exponential And Logarithmic Functions

Solve for x:

$log_{7}x=3$

Possible Answers:

x=343

x=21

x=2187

x=49

Correct answer:

x=343

Explanation:

In order to solve for x in a logarithmic function, we need to change it into exponential form. That looks as follows:

$log_{b}x=y\rightarrow b^{y}=x$

For this problem, manipulate the log and solve:

$log_{7}x=3$

$7^{3}=x$

x=343

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Example Question #71 : Exponential And Logarithmic Functions

What is the correct value of ? 3log_b (10)=1

Possible Answers:

log(300)

1000

300

$\frac{1}{1000}$

Correct answer:

1000

Explanation:

Divide by three on both sides.

$\frac{3log_b (10)}{3}=\frac{1}{3}$

$log_b (10)=\frac{1}{3}$

If we would recall 10^2 = 100 and $log_{10}(100) = 2$ , this indicates that:

$b^{\frac{1}{3}} = 10$

Cube both sides to isolate b.

$(b^{\frac{1}{3}}) ^3= 10^3$

The answer is: 1000

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Example Question #74 : Exponential And Logarithmic Functions

What is the value of 3log_4(64) ?

Possible Answers:

Correct answer:

Explanation:

The expression can be rewritten as:

$3log_4(64) = 3(\frac{log(64)}{log (4)}) = 3(\frac{log(4^3)}{log(4^1)}) =3(\frac{3log(4)}{log(4)})$

The answer is:

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Example Question #75 : Exponential And Logarithmic Functions

Solve this logarithm: $\log_{10} (5x-1)=\log_{10} (x+9)$

Possible Answers:

$x=\frac{1}{5}$

$x=\frac{2}{5}$

None of these

$x=\frac{10}{2}$

$x=\frac{5}{2}$

Correct answer:

$x=\frac{5}{2}$

Explanation:

By the one-to-one property of logarithms we are able to set 5x-1=x+9 and solve.

$5x-1=x+9\rightarrow 5x=x+10\rightarrow 4x=10\rightarrow \boldsymbol{x=\frac{5}{2}}$

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Example Question #21 : Solving Logarithmic Functions

Solve the logarithm: $-8-3\ln x=13$

Possible Answers:

$x\approx 3.12 \times 10^{4}$

$x\approx 3.12 \times 10^{-4}$

$x\approx 9.12 \times 10^{-4}$

$x\approx 2.19\times 10^{-4}$

$x\approx 1.92 \times 10^{-4}$

Correct answer:

$x\approx 9.12 \times 10^{-4}$

Explanation:

$-8-3\ln x=13$

add 8 to both sides:

$-3\ln x=21$

divide both sides by -3:

$\ln x=-7$

exponentiate both sides:

$e^{\ln x}=e^{-7}$

$x=\boldsymbol{e^{-7}\approx9.12\times 10^{-4}}$

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Example Question #1 : Applications

On the day of a child's birth, a sum of money is to be invested into a certificate of deposit (CD) that draws $6.2\%$ annual interest compounded continuously. The plan is for the value of the CD to be at least $\$20,000$ on the child's $18^{th}$ birthday.

If the amount of money invested is to be a multiple of $\$1,000$ , what is the minimum that should be invested initially, assuming that there are no further deposits or withdrawals?

Possible Answers:

$\$9,000$

$\$10,000$

$\$6,000$

$\$7,000$

$\$8,000$

Correct answer:

$\$7,000$

Explanation:

If we let be the initial amount invested and be the annual interest rate of the CD expressed as a decimal, then at the end of years, the amount of money that the CD will be worth can be determined by the formula

$A = Pe^{rt}$

Substitute A = 20,000 , r = 0.062 , t = 18 , and solve for .

$20,000 = Pe^{0.062 \cdot 18}$

$P = \frac{20,000 }{e^{0.062 \cdot 18}} \approx \frac{20,000 }{e^{1.116}} \approx \frac{20,000 }{3.0526} \approx 6,551$

The minimum principal to be invested initially is $6,551. However, since we are looking for the multiple of $1,000 that guarantees a minimum final balance of $20,000, we round up to the nearest such multiple, which is $7,000 - the correct response.

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Example Question #3311 : Algebra Ii

Twelve years ago, your grandma put money into a savings account for you that earns $7.5\%$ interest annually and is continuously compounded. How much money is currently in your account if she initially deposited $\$10,000$ and you have not taken any money out?

Possible Answers:

$8,103

$24,596

$81,030

$10,778

$21,170

Correct answer:

$24,596

Explanation:

1. Use $x=Pe^{rt}$ where is the current amount, is the interest rate, is the amount of time in years since the initial deposit, and is the amount initially deposited.

r=0.075

P=$10,000

t=12

2. Solve for

$x=Pe^{rt}$

$x=10,000e^{(0.075\cdot12)}$

x=24,596.03

You currently have $24,596 in your account.

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Example Question #76 : College Algebra

Jeffrey has won a lottery and has elected to take a $10,000 per month payment.

At the beginning of the year, Jeffrey deposits the first payment of $10,000 in an account that pays 7.6% interest annually, compounded continuously. At the very beginning of each month, he deposits another $10,000. How much will he have at the very end of the year?

Possible Answers:

$\$125,072.98$

$\$124,269.51$

$\$123,561.85$

$\$124,283.35$

$\$125.056.56$

Correct answer:

$\$125,072.98$

Explanation:

The continuous compound interest formula is

$A= Pe^{rt}$ ,

where is the amount of money in the account at the end of the period, is the principal at the beginning, is the annual interest rate in decimal form, and is the number of years over which the interest accumulates. Since Jeff deposits $1,000 per month, we apply this formula twelve times, with equal to the principal at the beginning of each successive month, r = 0.076 , and $t = \frac{1}{12} \approx 0.08333$ .

We can go ahead and calculate $e^{rt}$ , since and do not change:

$e^{rt}= e^{0.076 \cdot \frac{1}{12}} \approx e^{0.00633 } \approx 1.00635$

The formula can be rewritten as

A= P ( 1.006353)

At the beginning of January, Jeffrey deposits $10,000. At the end of January, there is

$A= \$10,000 ( 1.006353) = \$10,063.53$

in the account.

At the beginning of February, he again deposits $10,000, so there is now

$\$10,063.53+ \$10,000 = \$20,063.53$

in the account.

At the end of February, there is

$A= \$20,063.53 ( 1.006353) = \$ 20,191.01$

in the account.

Repeat addition of $10,000, then multiplication by 1.006353, ten more times to get the amount of money in the account at the end of December. This will be $125,072.98.

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College Algebra : Exponential and Logarithmic Functions

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #71 : Exponential And Logarithmic Functions

Example Question #72 : Exponential And Logarithmic Functions

Example Question #73 : Exponential And Logarithmic Functions

Example Question #71 : Exponential And Logarithmic Functions

Example Question #74 : Exponential And Logarithmic Functions

Example Question #75 : Exponential And Logarithmic Functions

Example Question #21 : Solving Logarithmic Functions

Example Question #1 : Applications

Example Question #3311 : Algebra Ii

Example Question #76 : College Algebra

All College Algebra Resources

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