Answer
$y\approx 3.2056$
Work Step by Step
Recall the definition of a logarithm (pg. 451):
$\log_{b}{x}=y$ iff $b^y=x$
Applying this definition to our equation (with $b=12, y=y-2, x=20$), we get:
$\log_{12}{20}=y-2$
Next, recall the change of base formula (pg. 464):
$\log_{b}{m}=\dfrac{\log_{c}{m}}{\log_{c}{b}}$
Applying this formula to our last equation, we get:
$\log_{12}{20}=y-2$
$\dfrac{\log_{10}{20}}{\log_{10}{12}}=y-2$
$\dfrac{\log_{10}{20}}{\log_{10}{12}}+2=y$
$y\approx 3.2056$
We confirm that the answer works:
$12^{3.2056-2}=20$
$12^{1.2056}=20$
$20=20$
