I got $\frac{6(\log 2)(\log 2)+5(\log 2)(\log 3)+6(\log 3)(\log 3)}{6(\log 2)(\log 2)+5(\log 2)(\log 3)+(\log 3)(\log 3)}$
Is it right saying it $\frac{1}{5(\log 3)(\log 3)}+1$
Hello, NameIsHidden!
I'm getting incorrect answers and I don't know why.
[I know ... I sound like a kid in Algebra I, but I can't help it.]
Divide [2] by [1]: .
Equate exponents: .
Solve the system: .
I checked with the solution, the result of the expression is $1$.
While taking Soroban's explanation, I see that he missed something.
Actually $a=-3$ or $a=3$ and $b=-2$ or $b=2$.(By substituting two equations in two different ways)
The final result is either $1$ or $11$.
Are there any other ways to compute it even accurate?