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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!

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1  
Logarithms are what you use to invert exponential functions. –  J. M. May 2 '12 at 12:54
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Wait...the question is to solve $h^{-1}(x)=2$. This is just $x=h(2)=9$, right? –  Simon Markett May 2 '12 at 12:59
    
If you are more interested in a general solution, then $3^x=e^{3ln(x)}$ might help –  Simon Markett May 2 '12 at 13:00
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As you can see from Simon's comment, the question isn't actually asking you to find the inverse function $h^{-1}$. If that's what you needed to do, you would indeed need logarithms, but as it is, you can treat $h^{-1}$ as a black box without worrying about what function it is, and just write $x=h(h^{-1}(x))=h(2)=3^2=9$. –  joriki May 2 '12 at 13:03

4 Answers 4

up vote 10 down vote accepted

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 $h^{-1}$.

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That's a great answer! Thanks. –  evmunro May 2 '12 at 13:18

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.

$$ h^{-1}(x) = 2 \rightarrow \log_3x = 2 \rightarrow x = 3^2 \rightarrow x = 9.$$

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(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!

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See Simon's comment under the question -- the detour through logarithms is unnecessary. –  joriki May 2 '12 at 13:05
    
I understand that. But I see 'enmunro' is attempting to "find the inverse of a function". I'll edit my answer to detail this. –  wangdw May 2 '12 at 13:10
    
I answered this question 16 mins ago... It's the earliest answer! –  wangdw May 2 '12 at 13:19
    
@DaweiWang: I see, sorry. –  Gigili May 2 '12 at 13:21
  1. First replace $h(x)$ with $y$ to get $y=3^x$.
  2. Swap $x, y$ to get $x=3^y$.
  3. Take, for example, $\log_3$ of both sides to get $\log_3x=y\log_33=y.$
  4. Replace $y$ with $h^{-1}(x)$.
  5. $h^{-1}(x)=\log_3(x)$.
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