Is $y = \log_e {(3 - x)}$ the inverse of $y = 3 - e^x$?

Question

I need to find the inverse function of $f(x) = 3 - e^x$.

Am I correct to assume it is $f^{-1}(x) = \log_e {(3 - x)}$?

Answer 1

Yes, you are correct. We get $$e^x=3-y$$ so $$x=\ln(3-y)$$ if $$3>y$$

Answer 2

Yes, your solution is correct. Here's how you find the inverse of an exponential function:

$$y=3-e^x\implies e^x=3-y.$$

A logarithmic function is the inverse of an exponential function by definition:

$$y=a^x\Longleftrightarrow x=\log_a{y}.$$

Therefore: $$3-y=e^x \Longleftrightarrow x=\ln{(3-y)}.$$

The only thing you need to do now is change the names of the variables.

You can also always check if whatever you have are inverse functions. For inverse functions, the following always holds true: $$f(f^{-1}(x))=x$$ and $$f^{-1}(f(x))=x.$$

Answer 3

Yes: If $f(x) = \ln (3 - x)$ then we will have:

$f(f^{-1}(x)) = x$

$\ln (3 - f^{-1}(x)) = x$

$e^{\ln (3 - f^{-1}(x))}= e^x$

$3-f^{-1}(x) = e^x$

$-f^{-1}(x) = e^x -3$

$f^{-1}(x) = 3 - e^x$

And we can verify that $\ln (3 - (3-e^x)) = \ln e^x = x$ and that $3-e^{\ln (3-x)} = 3- (3 -x) = x$.

Answer 4

Two function $f:X\to Y$ and $g:Y\to X$ are inverses of eachother if and only if $f(g(y))=y$ for every $y\in Y$ and $g(f(x))=x$ for every $x\in X$.

If $f(x)=\ln(3-x)$ then its (natural) domain is $(-\infty,3)$ and its range is $\mathbb R$.

So here it must be checked that:

• $\ln(3-(3-e^y)=y$ for every $y\in\mathbb R$.
• $3-e^{\ln(3-x)}=x$ for every $x\in(-\infty,3)$.

Now perform the checks yourself (and you will find that the functions are indeed inverses of each other).

It must not be neglected here that we are dealing with functions $(-\infty,3)\to\mathbb R$ and $\mathbb R\to(-\infty,3)$.

Answer 5

$$y=3-e^x\iff e^x=3-y\iff x=\ln(3-y),$$ so yes. These equations can only hold if $y<3$; there are no restrictions on $x$.

Answer 6

Set $f(x) = \ln (3-x)$ and $g(x) = 3 - e^{x}$.

If the two functions are inverses of each other, then $f(g(x)) = g(f(x)) = x$.

Well,

$$ f(g(x)) = \ln [3-(3 - e^{x})] = \ln (e^{x} = x$$

and

$$ g(f(x)) = 3 - e^{\ln (3-x)} = 3 - (3-x) = x.$$

Therefore, the two functions you cited are inverses of one another, where $\ln (3-x)$ is defined; that is, over $(-\infty, 3).$