Solve $(x^2+2x)^{x^2-3x}=12$

Question

Found this problem in a mathematics group site. I suspect it cannot be solved with pen and paper. Graphically $x=-2.438$ and $x=3.267$ but can it be solved algebraically? The answer was not posted on the site.

Answer 1

One can approximate the solution if required, rather say a pen and paper solution. Given that $(x^2+2x)^{x^2-3x}=12$

take log with base 10 on both sides, $(x^2-3x)\log(x^2+2x)=\log12$

LHS has domain $x\in(-\infty,-2)\cup(0,\infty)$, and LHS<0 for $x$ in interval $(0,3)$ . When $x>3$, the function has $f'(x)>0$, i.e. increases. Therefore, one solution is greater than 3. Testing $x=4$ in the original equation gives the result $331776$ which is extremely high, hence we can predict the solution lies is in interval $(3,3.5)$.

The second solution is smaller than zero. Note than LHS is not for $(-2,0)$ and shows similar behaviour as with x>3 in mirrored sense Testing for limit of 2 and 3 does the trick again.