# How do you solve log y = log (x-1) + 1?

y = $10 \left(x - 1\right)$, x > 1.
Working on common logarithm, y = 10^((log(x-1)+1).
${10}^{m + n} = {10}^{m} {10}^{n}$ and ${10}^{\log a} = a$.
$y = \left(x - 1\right) X {10}^{1} = 10 \left(x - 1\right)$
Here, $\log \left(x - 1\right)$ is defined for x > 1. So, the answer is suject yo x > 1..