Logarithms with Exponents

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Algebra II › Logarithms with Exponents

Questions 1 - 10

1

Which statement is true of

$\log_{7} \sqrt[N]{7}$

for all integers ?

$\log_{7} \sqrt[N]{7}= \frac{1}{N}$

$\log_{7} \sqrt[N]{7} = -N$

$\log_{7} \sqrt[N]{7} = N - 1$

$\log_{7} \sqrt[N]{7} = N - 7$

$\log_{7} \sqrt[N]{7}= \frac{7}{N}$

Explanation

Due to the following relationship:

$\sqrt[N]{7} = 7 ^{\frac{1}{N}}$ ; therefore, the expression

$\log_{7} \sqrt[N]{7}$

can be rewritten as

$\log_{7} \left ( 7 ^{\frac{1}{N}} \right )$

By definition,

$\log_{a} \left (a^{M} \right )= M$ .

Set and $M = \frac{1}{N}$ , and the equation above can be rewritten as

$\log_{7} \left ( 7 ^{\frac{1}{N}} \right )= \frac{1}{N}$ ,

or, substituting back,

$\log_{7} \sqrt[N]{7}= \frac{1}{N}$

2

Which statement is true of $\frac{1}{5} \log _{5}N$ for all positive values of ?

$\frac{1}{5} \log _{5}N = \log_{5} \sqrt[5]{N}$

$\frac{1}{5} \log _{5}N = \log_{5}\left ( \frac{1}{5} N \right )$

$\frac{1}{5} \log _{5}N = \log_{5} (N - 5)$

$\frac{1}{5} \log _{5}N = \log_{5} (N - 1)$

$\frac{1}{5} \log _{5}N = N$

Explanation

By the Logarithm of a Power Property, for all real , all ,

$b \log_{a} N = \log_{a}( N ^{b})$

Setting $b = \frac{1}{5} , a = 5$ , the above becomes

$\frac{1}{5} \log _{5}N = \log_{5} \left ( N^{\frac{1}{5}}\right )$

Since, for any for which the expressions are defined,

$N^{\frac{1}{b}} = \sqrt[b]{N}$ ,

setting , th equation becomes

$\frac{1}{5} \log _{5}N = \log_{5} \sqrt[5]{N}$ .

3

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

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

$\small b$

$\small a$

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

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

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

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 $\small b$

4

Simplify: $2e^{ln(2x+1)}$

$2^{2x+1}$

$x+\frac{1}{2}$

Explanation

According to log rules, when an exponential is raised to the power of a logarithm, the exponential and log will cancel out, leaving only the power.

Simplify the given expression.

$2e^{ln(2x+1)} = 2(2x+1)$

Distribute the integer to both terms of the binomial.

The answer is:

5

Simplify

$\log6^2$

$2\log6$

$\log3$

$\log12$

$(\log6)^2$

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.

$2 \log(6)$

6

Evaluate the following expression

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

$\small 9$

$\small 6$

$\small 3$

$\small \log (2000)$

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

Explanation

This is a simple exponent of a logarithmic answer.

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

$\small 3^{2}=9$

7

Evaluate the following expression

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

$\small 64$

$\small 48$

$\small 1024$

$\small 12$

$\small 4$

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

$\small 4^{3}=64$

8

Which of the following equations is valid?

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

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

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

none of the other answers are correct

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

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.

9

Use

$\log4=0.602$

and

$\log12=1.079$

Evaluate:

$\log64$

Explanation

Since the question gives,

$\log4=0.602$

and

$\log12=1.079$

To evaluate

$\log64$

manipulate the expression to use what is given.

$=\log4^3$

$=3\log4$

10

Evaluate the following expression

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

$\small 4$

$\small 27$

$\small 81$

$\small 2$

$\small 1/4$

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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