Finding the inverse of $h(x) = 3^x$
most of the time I know how to find the inverse of a function (make it equal $y$, solve for $x$ and then swap $x$ and $y$), but I have no idea how to do that for this one, so any help would be great: $h(x)=3^x$
From the question:
Solve the equation $h^{-1}(x)=2$
Thanks in advance for any help!
$\endgroup$ 46 Answers
$\begingroup$I think both existing answers are sort of missing the point of the exercise. You don't need to know anything about logarithms to do this exercise; all you need are the formal properties of inverse functions. The solution
$$x=h(h^{-1}(x))=h(2)=3^2=9$$
uses only the specific form of the function $h$ and the general formal properties of inverse functions, not the specific form of $h^{-1}$.
$\endgroup$ 1 $\begingroup$You know the process of finding the inverse, so let's go through it step by step.
First, replace $h(x)$ with $y$, $\quad y = 3^x$. Then, switch the $x$ and the $y$, $\quad x = 3^y$. Now, you have to solve for $y$ to find the inverse function. We can't take the $y$-th root of both sides, so in order to solve for $y$, we want to find the exponent that turns 3 into $x$. This is what's called a logarithm. By definition. $y = \log_ax$ if and only if $x = a^y$, that is, $\log_ax$ is the exponent that turns the base $a$ into $x$.
With this in mind, the inverse of $h(x) = 3^x$ would be $h^{-1}(x) = \log_3x$.
To solve the equation $h^{-1}(x) = 2$ we use the above definition.
$$ h^{-1}(x) = 2 \rightarrow \log_3x = 2 \rightarrow x = 3^2 \rightarrow x = 9.$$
$\endgroup$ $\begingroup$(I'd like to comment but I don't have that much reputation, sorry...)
You do need logarithms to invert this exponential function! The inverse will be
$$ h^{-1}(x)=\log_3x $$
Therefore, let $2=\log_3x$, we immediately see that $x=3^2=9$.
Additionally: Thanks to @joriki, I believe I need to make things clear. This exercise requires you to establish a better understanding of what an inverse to a function actually means. See joriki's answer!
$\endgroup$ 4 $\begingroup$- First replace $h(x)$ with $y$ to get $y=3^x$.
- Swap $x, y$ to get $x=3^y$.
- Take, for example, $\log_3$ of both sides to get $\log_3x=y\log_33=y.$
- Replace $y$ with $h^{-1}(x)$.
- $h^{-1}(x)=\log_3(x)$.
Denoting f{x) =y, just interchange x and y . It is that simple and also it is done!
$y=3^x $ becomes
$x=3^y , $ or $y_{inv} = \log_3(x) $
Now if $ \log_3(x) =2, x= 3^2 =9. $
You can graph $ f(x) $ and $ f^{-1}(x) $ to check that they indeed intersect on
the line $x=y,$ and nothing more needs to be done.
$\endgroup$ $\begingroup$The best way to solve the question is $h^{-1}(x) =2$ then $x=h (2)$
Now solve $h (2) =3^2=9$
So $x=9$
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