Convert Between Logarithmic and Exponential Practice

1. Rewrite the following equation in logarithmic form.

   $\quad 0.25=2^{ \Large{-2}}$

   Rewrite the following equation in exponential form.

   $\quad \log_{8}{\left(512\right)}=3$

2. The $3$ points plotted below are on the graph of $y=b^x$.

   Based only on these $3$ points, plot the $3$ corresponding points that must be on the graph of $y=\log_b{x}$.

   
   

3. Table I contains outputs of the function $f(x)=b^x$ for some $x$ values, and Table II contains outputs of the function $g(x)=\log_b(x)$ for some $x$ values. In both functions, $b$ is the same positive constant.

   Fill in the missing values in the tables. If necessary, round your answer to three decimal places.

   You do not need a calculator.

   Table I

   
   

   $x$	$0.631$	_____	$2$	$2.183$
   $f(x)=b^x$	$2$	$6$	$9$	$11$

   
   

   Table II

   $x$	$2$	$3$	$6$	_____
   $g(x)=\log_{b}(x)$	$0.631$	$1$	$1.631$	$2$

   
   

4. Rewrite the following equation in logarithmic form.

   $\quad 25=5^{ \Large{2}}$

   Rewrite the following equation in exponential form.

   $\quad \log_{32}{\left(16\right)}=\dfrac{4}{5}$
   

Source: Khan Academy, https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:logs/x2ec2f6f830c9fb89:log-intro/e/understanding-logs-as-inverse-exponentials
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1. 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:

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

   ---

2. 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.

   
   

   ---

3. 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

   
   

   $x$	$0.631$	$1.631$	$2$	$2.183$
   $f(x)=b^x$	$2$	$6$	$9$	$11$

   
   

   Table II

   $x$	$2$	$3$	$6$	$9$
   $g(x)=\log_{b}(x)$	$0.631$	$1$	$1.631$	$2$

   ---

4. 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:

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