## College Algebra (10th Edition)

$5\log_7{x}$
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: $=\log_7{x}(1+1+1+1+1) \\=\log_7{x}(5) \\=5\log_7{x}$