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Precalculus : Exponential and Logarithmic Functions

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Example Questions

Example Question #1 : Exponential And Logarithmic Functions

You are given that $\log_{10} x = A$ and $\log_{10} y = B$ .

Which of the following is equal to $100^{2A-B}$ ?

Possible Answers:

$\frac{x}{\sqrt{y}}$

$\frac{x^{4}}{y^{2}}$

$\frac{y^{2}}{x^{4}}$

$x^{4}y^{2}$

$\frac{2x}{y}$

Correct answer:

$\frac{x^{4}}{y^{2}}$

Explanation:

Since $\log_{10} x = A$ and $\log_{10} y = B$ , it follows that $10^{A} = x$ and $10^{B} = y$

$100^{2A-B} = \left( 10^{2} \right ) ^{2A-B} =10^{2 (2A-B)} =10^{4A-2B}=\frac{10^{4A}}{10^{2B}} =\frac{\left ( 10^{A}\right )^4}{\left (10^{B}\right )^2}=\frac{x^4}{y^2}$

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Example Question #2 : Logarithmic Functions

Solve.

$\ln(x^2-4)+\ln(3)=0$

Possible Answers:

$x=\pm\sqrt{\frac{4}{3}}$

None of the other answers

$x=\pm2$

x=4

$x=\pm\sqrt{\frac{13}{3}$

Correct answer:

$x=\pm\sqrt{\frac{13}{3}$

Explanation:

First we condense the LHS to $\ln(3x^2-12)$ using the properties of logarithms. Then we can eliminate the natural log on the LHS by raising both sides of the equation as exponents with base . This leaves us with 3x^2-12=1 since e^0 = 1 . Now we do some simple Algebra and obtain our answer.

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Example Question #111 : Pre Calculus

Solve.

$10^{2x}=64$

Possible Answers:

$\ln(8)$

$2\log(8)$

$\log(64)$

$\log(8)$

None of the other answers

Correct answer:

$\log(8)$

Explanation:

First take the logarithm of both sides. This yields $\log10^{2x}=\log(64)$ . Now using the properties of logarithms, we can bring the down in front of $\log(10)$ . Since $\log(10)=1$ . We are left with $2x = \log(64)$ . Dividing each side by we find $x = \frac{1}{2}\log(64) = \log(\sqrt{64})=\log(8)$ .

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Example Question #4 : Exponential And Logarithmic Functions

Evalute the equation e^2^x^+^1=100 . Use a calculator to approximate the answer to three decimal places.

Possible Answers:

4.605

1.803

1.021

2.281

0.500

Correct answer:

1.803

Explanation:

First, we rewrite the equation in logarithmic form to obtain $2x+1=\ln100$ . Simplifying, we get

2x = ln100 - 1

$x=\frac{ln100-1}{2}$

$x\approx1.803$

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Example Question #4 : Logarithmic Functions

Simplify

$\log_464$

Possible Answers:

Correct answer:

Explanation:

$\log_464\rightarrow \log_44^3\rightarrow 3$

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Example Question #1 : Exponential And Logarithmic Functions

Solve the following equation for :

$4^x-16^{x-2}=0$

Possible Answers:

Correct answer:

Explanation:

Logs must be used to solve this problem. Follow the steps below:

$4^x=16^{x-2}$

$4^x=4^{2(x-2)}$

$\log_{4}4^x=\log_{4}4^{2(x-2)}$

x=2(x-2)

x=2x-4

-x=-4

x=4

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Example Question #6 : Logarithmic Functions

Solve.

$\log_{2}(p-7)+\log_{2}(p)=3$

Possible Answers:

p=7,2

p=7,-2

p=-8,1

p=-8,-1

p=8

Correct answer:

p=8

Explanation:

First we condense the LHS to $\log_{2}(p^2-7p)$ . Now we can eliminate a log of base by making both sides of the equation exponents of base . This leaves us with p^2-7p=8 . Moving to the LHS we obtain a simple quadratic which we solve to obtain and for solutions. Finally, we eliminate as logarithms are undefined for negative numbers.

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Example Question #6 : Logarithmic Functions

Find the domain of the function f(x)= $\log_{4}(8 - 4x)$ .

Possible Answers:

(0,2)

$(-\infty,2)$

$[0,\infty )$

$(-\infty ,\infty )$

$(2,\infty )$

Correct answer:

$(-\infty,2)$

Explanation:

The logarithmic function is undefined when the inputs are negative or 0. Therefore the inputs of the logarithmic function must be positive. This means that the quantity 8-4x must be positive. After setting up the appropriate inequality, we have,

8-4x>0

-4x>-8

x<2

Therefore the domain of the function f(x)= $\log_{4}(8 - 4x)$ is the interval $(-\infty,2)$ .

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Example Question #2 : Logarithmic Functions

Use the logarithmic base change formula to convert $\log_38$ to a quotient of logarithms with a base of .

Possible Answers:

$\log_38=\frac{\log_48}{\log_43}$

$\log_38=\frac{\log_43}{\log_48}$

$\log_38=\log_43$

$\log_38=\log_4\frac{8}{3}$

$\log_38=\log_4\frac{3}{8}$

Correct answer:

$\log_38=\frac{\log_48}{\log_43}$

Explanation:

Here we employ the use of the logarithm base change formula

$\log_bx= \frac{\log_cx}{\log_cb}$ .

Therefore we get,

$\log_38=\frac{\log_48}{\log_43}$

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Example Question #6 : Logarithmic Functions

Find the domain of the function $f(x)=\log_{10}(1+\sqrt{-6-3x})$ .

Possible Answers:

$(-\infty,-2]$

$(2,\infty)$

$(-\infty,2]$

$(-\infty,-2)$

$(-2,\infty)$

Correct answer:

$(-\infty,-2]$

Explanation:

The logarithmic function is undefined when the inputs are negative or 0. Therefore the inputs of the logarithmic function must be positive. This means that the quantity $1+\sqrt{-6-3x$ must be positive. However, this quantity is always positive because the value under the radical cannot be negative. So we are only constrained by:

$\sqrt{-6-3x} \geq 0$

$-6-3x \geq0$

$-3x\geq6$

$x\leq-2$

Therefore the domain of the function $f(x)=\log_{10}(1+\sqrt{-6-3x})$ is the interval $(-\infty,-2]$ .

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Precalculus : Exponential and Logarithmic Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Exponential And Logarithmic Functions

Example Question #2 : Logarithmic Functions

Example Question #111 : Pre Calculus

Example Question #4 : Exponential And Logarithmic Functions

Example Question #4 : Logarithmic Functions

Example Question #1 : Exponential And Logarithmic Functions

Example Question #6 : Logarithmic Functions

Example Question #6 : Logarithmic Functions

Example Question #2 : Logarithmic Functions

Example Question #6 : Logarithmic Functions

All Precalculus Resources

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