Algebraic expression to the algebraic expressionth power equation

Question

How can you solve problems like $ x^{x-1}=7 $? More generally, how can you solve equations like $(ax+b)^{cx+d}=e$ , where $a,b,c,d,e$ are given?$($Give all the roots, including complex ones$)$

Answer 1

Interesting enough for me to post an answer is the fact that the problem for your first equation only lies in the fact that the exponent is given by $x-1$ and not by $x$. I will demnonstrate how to solve this equation for the latter case. First of all rewrite the $x^x$ term in terms of the exponential and do not forget assuming a complex valued logarithm to get

$$\begin{align*} x^x&=7\\ e^{x\log(x)+2\pi i n}&=7\\ x\log(x)+2\pi i n&=\log(7)\\ \log(x)e^{\log(x)}&=\log(7)-2\pi i n\\ \log(x)&=W(\log(7)-2\pi i n) \end{align*}$$

$$\therefore~x=e^{W(\log(7)-2\pi i n)}~~~n\in\mathbb Z$$

I have doubts that on can deduce a general formula for arbitrary $a,b,c,d,e$ $($just take $a=c=1$,$b=0$,$d=-1$ and $e=7$ to reproduce your first equation$)$. Anyway considering that $a=c$ and $b=d$ it is indeed possible since this is basically the same as $x^x$ and can be solved using the Lamber W-Function but note that you have to consider the different branches of this function with regard to the values of $a,b$ and $e$.

From hereon I have to admit that I have not enough experience with the Lambert W-Function to give an detailed outline of the different branches and why they are important.

Answer 2

$$x^{x-1}=e^{(x-1)\log x}\implies x^{x-1}=7\iff e^{(x-1)\log x}=e^{\log 7}\implies(x-1)\log x=\log7$$

and you'll probably need some trascendental function like Lambert function or stuff.