Answer
$1$
Work Step by Step
Recall the power property of logarithms (pg. 462):
$\log_b{m^n}=n\log_b{m}$
We rewrite the values in the form of $m^n$, so that we can apply
the power property:
$\log_{3}{27}-2\log_3{3}\\
=\log_{3}{3^3}-2\log_3{3}\\
=3\log_{3}{3}-2\log_3{3}\\
=(3\times1)-(2\times1)\\
=3-2\\
=1$
We also used the fact that $\log_{3}{3}=1$ (because $3^1=3$).
