Inverse of $4-\ln(x+2)$ How to find an inverse of function $4-\ln(x+2)$? I know that inverse of $\ln(x)$ equals $e^{x}$, then inverse of $-\ln(x)$ equals $e^{-x}$ but what to do with the other numbers (4 and 2)?
 A: If $f(x) = 4 - \log(x+2)$, then $4-f(x) = \log(x+2)$, so $\exp(4-f(x)) = x+2$, so $\exp(4-f(x)) - 2 = x$.
That is, the inverse function is $f^{-1}: f(x) \mapsto e^{4-f(x)}-2$. Alternatively stated, $f^{-1}(y) = e^{4-y} - 2$.
A: Let $f: x \mapsto 4 - \log (x+2) :\ ]-2, \infty[ \to \mathbb{R}\setminus \{ 4 \}$. We claim that $f$ is bijective.
To prove surjectiveness, let $y \in \mathbb{R}\setminus \{4 \}$. Since $4 - y = \log (x+2)$ for some $x > -2$ iff $x = e^{4-y} - 2$, by definition $f$ is surjective.
Let $x_{1},x_{2} > -2$ such that $x_{1} \neq x_{2}$. We claim that $f$ is injective. But if $f(x_{1}) = f(x_{2})$ then $\log(x_{1} + 2) = \log(x_{2}+2)$, so $x_{1} = x_{2}$, a contradiction. It follows that $f$ is injective. Therefore, by definition $f$ is bijective.
We claim that $f^{-1}: y \mapsto e^{4-y} - 2 :\ \mathbb{R}\setminus \{4 \} \to ]-2,\infty[$. But we have the identities $f\circ f^{-1}(y) = y$ and $f^{-1}\circ f(x) = x$ by construction, so we are done.  
A: Notice, we have $$y=f(x)=4-\ln(x+2)$$
$$\ln(x+2)=4-y$$
$$x+2=e^{4-y}$$
$$x=e^{4-y}-2$$
Now, for inverse function, interchanging $x$ & $y$, we get 
$$y=e^{4-x}-2$$
Hence, inverse function is  $$\color{red}{f^{-1}(x)}=\color{blue}{e^{4-x}-2}$$
