Inverse function of $2^{x(x-1)}$?

Question

I tried taking log to the base $2$ both sides and solving it using quadratic formula :

$$y = 2^{x(x-1)}$$

Taking log to the base $2$ both sides :

$$\log_2(y) = x(x-1)$$

$$x^2 - x -\log_2(y) = 0$$

Solving the above equation for $x$ :

$$x=\frac{1\pm\sqrt{1+4\log_2(y)}}2$$

However, the answer to this question is :

$$x = \frac{\log_2(y)}{\log_2(y) - 1}$$

I would appreciate if someone can answer this question for me. It has been bugging me for a while. Btw before posting this question here, I tried to find if someone has already asked this question but no one has.

Answer 1

It might be a typo in the book, indeed if we try to obtain the inverse function of $$y=2^{\frac x{x-1}}\,$$ we get the correct answer.

By taking the logarithm to the base $2$ both sides, we get that :

$$\log_2(y)=\dfrac x{x-1}$$

$$x\log_2(y)-\log_2(y)=x$$

and by solving the above equation for $x$ :

$$x\log_2(y)-x=\log_2(y)$$

$$x\big(\log_2(y)-1\big)=\log_2(y)\;\,.$$

Hence, the inverse function is :

$$x=\frac{\log_2(y)}{\log_2(y)-1}\;\,.$$