Answer
$2$ + $\log_{7}$ ${x}$ + $\log_{7}$ ${y}$ + $\log_{7}$ ${z}$
Work Step by Step
Use the Product Property of Logarithms. According to this property, $\log_b$ ${mn}$ = $\log_b$ ${m}$ + $\log_b$ ${n}$. Thus, the given expression is equivalent to:
$\log_{7}$ ${49}$ + $\log_{7}$ ${x}$ + $\log_{7}$ ${y}$ + $\log_{7}$ ${z}$
Rewrite $49$ as a power of $7$:
$\log_{7}$ ${7^2}$ + $\log_{7}$ ${x}$ + $\log_{7}$ ${y}$ + $\log_{7}$ ${z}$
Use the Power Property of Logarithms to rewrite this expression. The property states that $\log_b$ ${m^n}$ = $n$ $\log_b$ ${m}$. Thus, the expression above is equivalent to:
$2$ $\log_{7}$ ${7}$ + $\log_{7}$ ${x}$ + $\log_{7}$ ${y}$ + $\log_{7}$ ${z}$
Use the logarithmic identity $\log_{b}$ ${b}$ = $1$ to obtain:
$2(1)$ + $\log_{7}$ ${x}$ + $\log_{7}$ ${y}$ + $\log_{7}$ ${z}$
Simplify:
$2$ + $\log_{7}$ ${x}$ + $\log_{7}$ ${y}$ + $\log_{7}$ ${z}$
