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Algebra II : Logarithms with Exponents

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Example Questions

Example Question #1 : Logarithms With Exponents

In this question we will use the notation $\small \log a$ to represent the base 10 or common logarithm, i.e. $\small \log a \equiv \log_{10}a$ .

Find if 3^x=12 .

Possible Answers:

$\small x \approx 4.000$

$\small x \approx 2.9891$

$\small x \approx 2.2620$

$\small x \approx 2.3116$

$\small x \approx 0.4771$

Correct answer:

$\small x \approx 2.2620$

Explanation:

We can use the Property of Equality for Logarithmic Functions to take the logarithm of both sides:

$\small \log 3^x = \log 12$

Use the Power Property of Logarithms:

$\small x \log 3 = \log 12$

Divide each side by $\log3$ :

$\small x=\frac{\log 12}{\log 3}$

Use a calculator to get:

$\small x \approx \frac{1.0792}{0.4771}$

$\small x \approx 2.2620$

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Example Question #1 : Logarithms With Exponents

Simplify

$\log6^2$

Possible Answers:

$\log12$

$2\log6$

$\log3$

$(\log6)^2$

Correct answer:

$2\log6$

Explanation:

Using Rules of Logarithm recall:

$\log a^b = b\log a$ .

Thus, in this situation we bring the 2 in front and we get our solution.

$2 \log(6)$

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Example Question #1 : Logarithms With Exponents

Simplify the following equation.

$(e^{(\ln2+\ln93-\ln6)}e^{-1})^2$

Possible Answers:

$\frac{961}{e^2}$

$\frac{31}{e}$

$\frac{31}{e^{-2}}$

31e

961e^2

Correct answer:

$\frac{961}{e^2}$

Explanation:

We can simplify the natural log exponents by using the following rules for naturla log.

$\ln A+\ln B=\ln AB$

$\ln A - \ln B=\ln\frac{A}{B}$

Using these rules, we can perform the following steps.

$(e^{(\ln2+\ln93-\ln6)}e^{-1})^2$

$(e^{(\ln(2(93))-\ln6)}e^{-1})^2$

$(e^{(\ln(\frac{2(93)}{6})}e^{-1})^2$

$(e^{\ln 31}e^{-1})^2$

Knowing that the e cancels the exponential natural log, we can cancel the first e.

$(31e^{-1})^2$

Distribute the square into the parentheses and calculate.

$(31)^2(e^{-1})^2=961e^{-2}$

Remember that a negative exponent is equivalent to a quotient. Write it as a quotient and then you're finished.

$961e^{-2}=\frac{961}{e^2}$

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Example Question #1 : Logarithms With Exponents

Evaluate the following expression

$\small \log(1000^{2})$

Possible Answers:

$\small 9$

$\small \frac{1}{9}$

$\small 3$

$\small 6$

$\small \frac{\log (2)}{\log (10)}$

Correct answer:

$\small 6$

Explanation:

Since the exponent is inside the parentheses, you must take the square of 1000 before finding the logarithim. Therefore

$\small \log(1000^{2})=\log (1000000)=6$

because $\small 10^{6}=1,000,000$

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Example Question #1 : Logarithms With Exponents

Evaluate the following expression

$\small \log _{3}(9^{2})$

Possible Answers:

$\small 27$

$\small 81$

$\small 1/4$

$\small 2$

$\small 4$

Correct answer:

$\small 4$

Explanation:

Since the exponent is inside the parentheses, you must determine the value of the exponential expression first.

$\small 9^{2}=81$

then you solve the logarithm

$\small \log _{3}(81)=4$ because $\small 3^{4}=81$

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Example Question #4 : Logarithms With Exponents

Evaluate the following for all integers of $\small a$ and $\small b$

$\small a^{\log _{a}(b)}$

Possible Answers:

$\small b$

$\small a$

$\small \frac{a}{b}$

$\small \log_{a} (b)$

$\small \log _{b}(a)$

Correct answer:

$\small b$

Explanation:

$\small \log _{a}(b)$ gives us the exponent to which $\small a$ must be raised to yield $\small b$

When $\small a$ is actually raised to that number in the equation given, the answer must be $\small b$

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Example Question #5 : Logarithms With Exponents

Evaluate the following expression

$\left ( \log1000\right)^{2}=$

Possible Answers:

$\small 9$

$\small \log (2000)$

$\small 6$

$\small 3$

$\small \frac{1}{1000}$

Correct answer:

$\small 9$

Explanation:

This is a simple exponent of a logarithmic answer.

$\small \log(1000)=3$ because $\small 10^{3}=1000$

$\small 3^{2}=9$

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Example Question #6 : Logarithms With Exponents

Evaluate the following expression

$\left( \log _{2}16 \right)^{3}$

Possible Answers:

$\small 48$

$\small 12$

$\small 4$

$\small 1024$

$\small 64$

Correct answer:

$\small 64$

Explanation:

This is a two step problem. First find the log base 2 of 16

$\small \log _{2}(16)=4$ because $\small 2^{4}=16$

then

$\small 4^{3}=64$

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Example Question #131 : Logarithms

Which of the following equations is valid?

Possible Answers:

$\small \log _{a}(a^{b})=b$

$\small a^{\log _{a}(b)}=a$

none of the other answers are correct

$\small \log_{2b}(a)= \log_{b}(a)^{2}$

$\small \log (a)^{2}=\log2a$

Correct answer:

$\small \log _{a}(a^{b})=b$

Explanation:

Since a logarithm answers the question of which exponent to raise the base to receive the number in parentheses, if the number in parentheses is the base raised to an exponent, the exponent must be the answer.

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Example Question #3 : Logarithms With Exponents

Rewrite the following logarithmic expression into expanded form (that is, using addition and/or subtraction):

$\small \ln{(AB)}^{4}$

Possible Answers:

$\small \ln{A^{4}}+\ln{B^4}$

$\small 4\ln{A}+4\ln{B}$

$\small 4\ln{A}+\ln{B}$

$\small 4\ln{AB}$

Correct answer:

$\small 4\ln{A}+4\ln{B}$

Explanation:

Before we do anything, the exponent of 4 must be moved to the front of the expression, as the rules of logarithms dictate. We end up with $\small 4\ln{AB}$ . Remember that a product inside of a logarithm can be rewritten as a sum: $\small 4(\ln{A}+\ln{B})$ . Distributing, we get $\small 4\ln{A}+4\ln{B}$ .

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Algebra II : Logarithms with Exponents

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : Logarithms With Exponents

Example Question #1 : Logarithms With Exponents

Example Question #1 : Logarithms With Exponents

Example Question #1 : Logarithms With Exponents

Example Question #1 : Logarithms With Exponents

Example Question #4 : Logarithms With Exponents

Example Question #5 : Logarithms With Exponents

Example Question #6 : Logarithms With Exponents

Example Question #131 : Logarithms

Example Question #3 : Logarithms With Exponents

All Algebra II Resources

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