# How do you solve 5^x = 25^(x-1) ?

${25}^{b} = {\left({5}^{2}\right)}^{b} = {5}^{2 b}$ This is a basic rule to be memorised.
${5}^{x} = {25}^{x - 1}$
${5}^{x} = {\left({5}^{2}\right)}^{x - 1} \implies {5}^{x} = {5}^{2 \cdot \left(x - 1\right)} \implies {5}^{x} = {5}^{2 x - 2}$ Since bases (5 values) are same. We can write;
$x = 2 x - 2 \implies \underline{x = 2}$