# How do you solve log_b121=2?

$\pm 11$, see explanation.
if you have ${a}^{b} = c$, logarithms go ${\log}_{a} c = b$. That's just a thing that you need to know, I'm not quite sure how to explain that, but if we apply it to the logarithm in the problem...
${\log}_{b} 121 = 2 \to {b}^{2} = 121$.
And from here you just take the square root of both sides and get $\pm 11$. It's plus or minus because ${11}^{2}$ and ${\left(- 11\right)}^{2}$ equal 121. (Negative multiplied by a negative equals a positive).