# How do you solve 9^(x-1)= 5^(2x)?

Mar 2, 2018

$x = - 2.150$

#### Explanation:

Start by taking the logarithm of both sides, in order to bring the exponents down.

${\log}_{10} \left({9}^{x - 1}\right) = {\log}_{10} \left({5}^{2 x}\right)$

One of the log rules is this:

${\log}_{a} \left({x}^{n}\right) = n {\log}_{a} \left(x\right)$

So:

$\left(x - 1\right) {\log}_{10} \left(9\right) = 2 x {\log}_{10} \left(5\right)$

Evaluate the logarithm and simplify:

$\left(x - 1\right) \left(0.9542\right) = 2 x \left(0.6990\right)$

$0.9542 x - 0.9542 = 1.398 x$

$- 0.9542 = 0.4438 x$

$x = - 2.150$