Convert Between Logarithmic and Exponential Practice

The inverse relationship of exponents and logarithms
For $m>0$ and $b>0, b\neq 1$, we have the following relationship:

$\qquad { b^{ {\Large{ q}}}}=m\quad$ if and only if $\quad \log_{ b }{m}= q$

Converting the exponential equation

So $\, {{2}^{ { \Large{-2}}}}={0.25}\,$ implies that $\,\log_{ {2}}({ {0.25}})= {-2}$.

Converting the logarithmic equation

Similarly $\, \log_{ 8}({{512}})= {3}\,$ implies that $\, 8^{ { \Large{3}}}={512}$.

The logarithmic form of $\,0.25=2^{ \Large{-2}}$ is:

$\log_2{{(0.25)}}={-2}$

The exponential form of $\log_{8}{\left(512\right)}=3$ is:

$\,8^{{ \Large{3}}}={512}$

Let's consider the point on $y = { b}^ x$ with coordinates $(1, 4)$.

Since $y = \log_{ b}{ x}$ is the inverse of $y={ b}^ x$, the point $( 4,1 )$ is on the graph of $y = \log_{b}{x}$.

In general, if $(p, q)$ is on $y={ b}^ x$, then $( q,p )$ is on $y = \log_{ b}{x}$.

For each point on $y=b^x$, we just switch the order of its coordinates to get a point on $y=\log_b{x}$.

So, $y=\log_b{x}$ also has points with coordinates $(1, 0)$ and $(16, 2)$.

Given the points that we know are on ${y=b^x}$, the graph below shows the $3$ points that must be on ${y=\log_b{x}}$.

The original $3$ points are also plotted for reference.

The inverse relationship of exponents and logarithms

By definition, we know that $f(x)=b^x$ and $g(x)=\log_b x$ are inverse functions.

Therefore, if $(p,q)$ satisfies function $f$, then we know that $(q,p)$ must satisfy function $g$.

Filling table I

From the second table, we see that $(6,1.631)$ satisfies function $g$, and so $\log_b{6}=1.631$.

This also implies that $b^{1.631}=6$ , and so $(1.631,6)$ satisfies function $f$. 

Filling table II

From the first table, we see that $(2,9)$ satisfies function $f$, and so $b^{2}=9$.

This also implies that $\log_b{9}=2$, and so $(9,2)$ satisfies function $g$. 

Here are the complete tables:

Table I

Table II

The inverse relationship of exponents and logarithms
For $m>0$ and $b>0, b\neq 1$, we have the following relationship:

$\qquad { b^{ {\Large{ q}}}}=m\quad$ if and only if  $\quad \log_{ b }{m}= q$

Converting the exponential equation

So $\, {{5}^{ { \Large{2}}}}={25}\,$ implies that $\,\log_{ {5}}({{25}})= {2}$.

Converting the logarithmic equation

Similarly $\, \log_{ {32}}\left({{16}}\right)= {\dfrac{4}{5}}\,$ implies that  $\, {32}^{ { \Large{\frac{4}{5}}}}={16}$.

The logarithmic form of $25=5^{ \Large{2}}$ is:

$\log_5{{(25)}}={2}$

The exponential form of $\log_{32}{\left(16\right)}=\dfrac{4}{5}$ is:

$\,32^{{ \Large{\frac45}}}={16}$