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Question
what is $log_{5}(4 cdot 7)+log_{5}2$ written as a single log?$log_{5}21$$log_{5}26$$log_{5}30$$log_{5}56$
Step1: Simplify inside first log
$\log_5(4 \cdot 7) = \log_5 28$
Step2: Apply log product rule
$\log_5 28 + \log_5 2 = \log_5(28 \cdot 2)$
Step3: Calculate the product
$\log_5(28 \cdot 2) = \log_5 56$
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$\log_5 56$ (corresponding to the last option)