Solve in $\mathbb{R}$ the following equation: $ (3^x+2)^{\log_5(3)} + 2 = (5^x-2)^{\log_3(5)} $
My approach: $ (3^x+2)^{\log_5(3)} + 2 = (5^x-2)^{\log_3(5)} = t $
After some simplification, we get: $ 5^x = 3^{\log_5(t)} + 2 \quad \text{and} \quad 3^x = 5^{\log_3(t-2)} - 2 $
Adding these two, we get: $ 3^x + 5^x = 3^{\log_5(t)} + 5^{\log_3(t-2)} $
I tried considering a function, but we have $t$ and $t-2$ which are different. Any help would be appreciated.
We have an expression of the form $A^{\log_5(3)}+2=B^{\log_3(5)}$ which implies that the fractional parts of exponentials are equal, i.e.$\{A^{\log_5(3)}\}=\{B^{\log_3(5)}\}$.This is rare enough excepting when both are zero.
Remember that $\boxed{a^{\log_a(b)}=b}$ we look at the equations $5^x-2=3$ and $3^x+2=5$ and apply the boxed formula so we get $3+2=5$.
The solution is $x=1$