Question

expand the following logarithmic expression into a sum or difference of logs. \\(\log_{19}\left(z^{10}w^{130}\
ight)\\) \\(\bigcirc\\) \\(\frac{10\log_{19}(z)}{130}\log_{19}(w)\\) \\(\bigcirc\\) \\(10\log_{19}(z) + 130\log_{19}(w)\\) \\(\bigcirc\\) \\(10 - \log_{19}(z) + 130\log_{19}(w)\\) \\(\bigcirc\\) \\(\frac{10}{\log_{19}(z)} - 130\log_{19}(w)\\) \\(\bigcirc\\) none of the above

Explanation:

Step1: Apply Product Rule of Logs

The product rule of logarithms states that $\log_b(MN) = \log_b(M) + \log_b(N)$. For $\log_{19}(z^{10}w^{130})$, we can split it as $\log_{19}(z^{10}) + \log_{19}(w^{130})$.

Step2: Apply Power Rule of Logs

The power rule of logarithms states that $\log_b(M^n) = n\log_b(M)$. Applying this to each term: $\log_{19}(z^{10}) = 10\log_{19}(z)$ and $\log_{19}(w^{130}) = 130\log_{19}(w)$.

Step3: Combine the Terms

Combining the results from Step 2, we get $10\log_{19}(z) + 130\log_{19}(w)$.

Answer:

B. $10 \log_{19}(z) + 130 \log_{19}(w)$