# 004 Sample Final A, Problem 19

Solve for x:      ${\displaystyle \log _{6}{\frac {1}{36}}=x}$

Foundations
How do we remove logs from an equation?
The definition of the logarithm tells us that if ${\displaystyle \log _{6}(x)=y}$, then ${\displaystyle 6^{y}=x}$.
By the definition of the logarithm, ${\displaystyle \log _{6}{\frac {1}{36}}=x}$ means ${\displaystyle 6^{x}={\frac {1}{36}}}$.
Now, we can solve for ${\displaystyle x}$. Since ${\displaystyle 6^{x}={\frac {1}{36}}}$,
we must have ${\displaystyle x=-2}$.
${\displaystyle x=-2}$