Exercise 70 Page 316

Let's start by recalling the definition of a logarithm. The above means that $c$ is the exponent to which the base $b$ must be raised to obtain $a.$ If we want an output of a logarithmic function to be $\text{-}4,$ we need to find two numbers $a$ and $b$ such that $a=b^{\text{-}4}.$ There are infinitely many numbers that satisfy this. One possible pair is $a=16$ and $b=\frac{1}{2}.$ A logarithmic function has the form $f(x)={\log }_{b}x.$ Therefore, to write a logarithmic function with $\text{-}4$ as one of its outputs, we can substitute $f(x)$ for $\text{-}4$ and $x$ for $16.$ For the function above, the output for $x=16$ is $\text{-}4.$ To graph the logarithmic function, we will first find its inverse. To do so, we will first replace $f(x)$ with $y,$ exchange the $x\text{-}$ and $y\text{-}$variables, and solve for $y.$ Let's finally isolate $y$ by applying the definition of a logarithm. The inverse function is $f^{\text{-}1}(x)={\left(\frac{1}{2}\right)}^x.$ Let's make a table of values to graph it.

$x$	${\left(\frac{1}{2}\right)}^x$	$f^{\text{-}1}(x)={\left(\frac{1}{2}\right)}^x$
$\text{-}4$	${\left(\frac{1}{2}\right)}^{\text{-}4}$	$16$
$\text{-}2$	${\left(\frac{1}{2}\right)}^{\text{-}2}$	$4$
$0$	${\left(\frac{1}{2}\right)}^0$	$1$
$1$	${\left(\frac{1}{2}\right)}^1$	$\frac{1}{2}$
$2$	${\left(\frac{1}{2}\right)}^2$	$\frac{1}{4}$

By plotting and connecting the ordered pairs, we can graph $f^{\text{-}1}(x).$ Then, we can reflect it in the line $y=x$ to have the graph of $f(x).$

As we can see, $f(x)={\log }_{1/2}x$ has an output of $\text{-}4$ at $x=16.$