Question

what is \\(\log_{5}(4 \cdot 7) + \log_{5}2\\) written as a single log?
\\(\circ\\) \\(\log_{5}21\\)
\\(\circ\\) \\(\log_{5}26\\)
\\(\circ\\) \\(\log_{5}30\\)
\\(\circ\\) \\(\log_{5}56\\)

Explanation:

Step1: Simplify the first log's argument

First, calculate the product inside the first logarithm: \(4 \cdot 7 = 28\), so the expression becomes \(\log_{5}28 + \log_{5}2\).

Step2: Apply the logarithm product rule

The logarithm product rule states that \(\log_{b}M + \log_{b}N=\log_{b}(M \cdot N)\). Here, \(b = 5\), \(M = 28\), and \(N = 2\). So we multiply the arguments: \(28 \cdot 2 = 56\). Thus, the expression simplifies to \(\log_{5}56\).

Answer:

\(\log_{5}56\) (corresponding to the option: \(\boldsymbol{\log_{5}56}\))