When the base $\displaystyle e$ is raised to a certain power, taking the natural log of this whole term will eliminate the exponential and the power can be pulled out as the coefficient.

$\displaystyle 2ln(e^3) = 2(3) ln(e^1) = 2(3)= 6$

The answer is:  $\displaystyle 6$

In order to solve this log, we will need to write 125 in terms of one fifth to a certain power.

Rewrite 125 as an exponent of one-fifth.

$\displaystyle 125 = (\frac{1}{5})^{-3}$

$\displaystyle log_{\frac{1}{5}} 125=log_{\frac{1}{5}}(\frac{1}{5})^{-3}$

According to the log rule,  $\displaystyle log_{x}x^y=y$, the bases will cancel, leaving just the exponent.

The answer is:  $\displaystyle -3$

Notice that the term in the log can be rewritten as a base raised to a certain power.

Rewrite the number in terms of base five.

$\displaystyle 2log(125)=2log(5^3)$

According to log rules, the exponent can be dropped as a coefficient in front of the log.

$\displaystyle 2\cdot 3log(5) = 6log(5)$

The answer is:  $\displaystyle 6log(5)$

Evaluate the log using the following property:

$\displaystyle log_xx^y = y$

The log based and the base of the term will simplify.

The expression becomes: 

$\displaystyle 3log_{5}(5^{10}) = 3(10) = 30$

The answer is:  $\displaystyle 30$

By definition, $\displaystyle \log_{7} 700 = 2$ if and only if $\displaystyle 7^{2} = 700$. However, 

$\displaystyle 7^{2} = 7 \times 7 = 49 \ne 700$,

making this false.

By definition, $\displaystyle \log_{16} 8 = N$ if and only if $\displaystyle 16^{N} = 8$.

8 and 16 are both powers of 2; specifically, $\displaystyle 8= 2^{3} , 16 = 2^{4}$. The latter equation can be rewritten as

$\displaystyle \left (2 ^{4} \right ) ^{N} = 2^{3}$

By the Power of a Power Property, the equation becomes

$\displaystyle 2 ^{4\cdot N} = 2^{3}$

or

$\displaystyle 2 ^{4N} = 2^{3}$

It follows that

$\displaystyle 4N = 3$,

and

$\displaystyle N = \frac{3}{4}$,

the correct response.

By definition, $\displaystyle \log_{a}N = M$ if and only if $\displaystyle a^{M}= N$. Set $\displaystyle a= 7, M = 1.8$, and

$\displaystyle \log_{7} N = 1.8$

if and only if

$\displaystyle N= 7^{1.8}$

Through calculation, we see that

$\displaystyle N \approx 33.2$.

The question asks for the value of the "base 0 logarithm" of 0. However, this is not defined, as a logarithm can only have as its base a positive number not equal to 1. 

By definition of a logarithm,

$\displaystyle \log_{8} M = N$ 

if and only if 

$\displaystyle 8^{N } = M$

Take the $\displaystyle N$th root of both sides, or, equivalently, raise both sides to the power of $\displaystyle \frac{1}{N}$, and apply the Power of a Power Property:

$\displaystyle \left (8^{N } \right ) ^{\frac{1}{N}} = M ^{\frac{1}{N}}$

$\displaystyle 8^{N \cdot \frac{1}{N} } = M ^{\frac{1}{N}}$

$\displaystyle 8^{1 } = M ^{\frac{1}{N}}$

or

$\displaystyle M ^{\frac{1}{N}} = 8$

By definition, it follows that $\displaystyle \log_{M} 8 = \frac{1}{N}$, so the statement is true.

By definition,

$\displaystyle \log_{N }10 = M$

If and only if

$\displaystyle N^{M}= 10$

Square both sides, and apply the Power of a Power Property to the left expression:

$\displaystyle (N^{M})^{2}= 10^{2}$

$\displaystyle N^{M\cdot 2}= 100$

$\displaystyle N^{2M}= 100$

It follows that for all positive $\displaystyle N$ not equal to 1,

$\displaystyle \log_{N }100 = 2M$ 

for all $\displaystyle N$.