Answer
$5\log_7{x}$
Work Step by Step
RECALL:
(1) $\log_a{(MN)} = \log_a{M} + \log_a{N}$
(2) $\log_a{\left(\dfrac{M}{N}\right)} = \log_a{M} - \log_a{N}$
(3) $\log_a{a} = 1$
Note that $x^5=x(x)(x)(x)(x)$. So the given expression is equivalent to:
$=\log_7{\left(x\cdot x\cdot x\cdot x\cdot x\right)}$
Using rule (1) above gives:
$=\log_7{x} + \log_7{x}+ \log_7{x}+ \log_7{x}+ \log_7{x}$
Factor out $\log_7{x}$ to obtain:
$=\log_7{x}(1+1+1+1+1)
\\=\log_7{x}(5)
\\=5\log_7{x}$
