Convert the following logarithmic equation to an exponential equation:

\(\displaystyle log_{13}b=147\)

Recall the following:

This

\(\displaystyle log_b(a)=c\)

Can be rewritten as

\(\displaystyle b^c=a\)

So, our given logarithm

\(\displaystyle log_{13}b=147\)

Can be rewritten as

\(\displaystyle 13^{147}=b\)

Fortunately we don't need to expand, because this woud be a very large number!

Rewrite the follwing equation as a logarithm:

\(\displaystyle 14^b=977\)

To complete this problem, recall the following relationship:

\(\displaystyle log_ba=c\) can be rewritten as  \(\displaystyle b^c=a\)

So, this:

\(\displaystyle 14^b=977\)

Is the same thing as this:

\(\displaystyle log_{14}977=b\)

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.