How to find inverse function for following logarithm function?

Question

I am kind stuck with finding the inverse function of following function.

$$f:x\mapsto\log_2(x^2-3x-4)$$

Let $y =f(x)=\log_2 (x^2 - 3x-4)$. After I simplify the function, I end up with: $2^x = y^2 - 3y -4$. But literally what is the next step to do, but the answer surprisingly looks like this:

$$\begin{split} f^{-1}(x)&=\frac{3+\sqrt{25+2^{x+2}}}{2}\quad\text{if }x>4\\ f^{-1}(x)&=\frac{3-\sqrt{25+2^{x+2}}}{2}\quad\text{if }x<-1\\ \end{split}$$

which seems like making quadratic formula, but have no clue on how the answer being derived.

Any help will be really appreciated.

Answer 1

The folloiwng can be tried always, though not always you get a nice output:

$$y=\log_2(x^2-3x-4)\implies2^y=x^2-3x-4\implies x^2-3x-(4+2^y)=0\;(*)$$

After we wrote the function as $y\;$ function of $\,x\;$ , we try the other way around: $\;x\;$ as function of $\;y\;$ . The above quadratic's discriminant:

$$\Delta=9+4(4+2^y)=25-4\cdot2^y=25+2^{y+2}$$

and the solutions are

$$x_{1,2}=\frac{3\pm\sqrt{25+2^{y+2}}}2\;\ldots$$

Now just interchange $\;x\leftrightarrow y\;$ to write the inverse in the usual way, and we're done.

Answer 2

Since $x^2-3x-4=(x+1)(x-4),$ the domain of $f$ is $${\rm dom}(f)=\{x\in\Bbb R\mid x^2-3x-4>0\}=(-\infty,-1)\cup(4,+\infty).$$ The answer you reproduced from your textbook is doubly wrong because

$f:{\rm dom}(f)\to\Bbb R$ (plot) is not injective, and is surjective.

For every $y\in\Bbb R$ (so: not only for $y>4$ or $<-1$), there are exactly two values of $x\in{\rm dom}(f)$ such that $y=f(x)$ (not only one), namely: the two solutions $x_\pm$ of the quadratic equation $$x^2-3x-(4+2^y)=0,$$ i.e. $$x_\pm=\frac{3\pm\sqrt\Delta}2,\text{ where }\Delta=(-3)^2+4(4+2^y)=25+2^{y+2}.$$ Note that these two values $x_{\pm}$ automatically belong to ${\rm dom}(f),$ since $x_\pm^2-3x_\pm-4=2^y>0.$