# How do you solve 25^ { x ^ { 2} - x } = 1?

Mar 25, 2017

${x}_{1} = 0$

${x}_{2} = 1$

#### Explanation:

Using the law of exponents, we know that the only exponent of $25$ which yields an answer of $1$ is $0$. $\left({\log}_{25} 1 = 0\right)$

Thus, ${x}^{2} - x = 0$

$x \left(x - 1\right) = 0$

$x = 0$

$x - 1 = 0$

$x = 1$