Question

Solve the equation below. Give the solution as an integer or reduced fraction.
log5 (x-7)= 2

Answer 1

`\begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{ \textit{we'll use this one} }{a^(log_a (x))=x} \end{array} \\\\[-0.35em] ~\dotfill\\\\ \log_5(x-7)=2\implies 5^(\log_5(x-7))=5^2\implies x-7=5^2 \\\\\\ x-7=25\implies x=32`