Answer
TRUE
Work Step by Step
Using the properties of logarithms and the laws of exponents, the left-hand expression, $
\log_3 49+\log_3 49^{-1}
$, is equivalent to
\begin{align*}
&
\log_3 \left(49\cdot49^{-1}\right)
&(\text{use }\log_b (xy)=\log_b x+\log_b y)
\\\\&=
\log_3 49^{1+(-1)}
&(\text{use }a^m\cdot a^n=a^{m+n})
\\&=
\log_3 49^{0}
\\&=
\log_3 1
&(\text{use }a^0=1)
\\&=
0
&(\text{use }\log_b 1=0)
.\end{align*}
Thus, the left-hand expression is equivalent to the given right-hand expression. Hence, the given statement is true.