Question

complete the steps to evaluate \\(\log_{7}98\\), given \\(\log_{7}2 \approx 0.356\\).
how can you rewrite \\(\log_{7}98\\) using the product property?
\\(\circ\\) \\(7\log2 + 7\log49\\)
\\(\circledcirc\\) \\(\log7 + \log2 + \log49\\)
\\(\circ\\) \\(\log_{7}2 + \log_{7}49\\)
done

Explanation:

The product property of logarithms states that \(\log_b(xy)=\log_b x+\log_b y\) for positive \(b
eq1\), \(x\), and \(y\). First, factorize \(98\): \(98 = 2\times49\). Now, apply the product property to \(\log_7 98=\log_7(2\times49)\). Using the property, this becomes \(\log_7 2+\log_7 49\) (since the base \(b = 7\) is consistent for both terms). The other options are incorrect: the first option incorrectly applies coefficients and bases, and the second option incorrectly introduces \(\log 7\) (the base is part of the logarithm notation, not a separate log term here).

Answer:

\(\boldsymbol{\log_7 2+\log_7 49}\) (the third option: \(\log_7 2+\log_7 49\))