what is $4log_{\frac{1}{2}}w+(2log_{\frac{1}{2}}u - 3log_{\frac{1}{2}}v)$ written as a single…

what is $4log_{\frac{1}{2}}w+(2log_{\frac{1}{2}}u - 3log_{\frac{1}{2}}v)$ written as a single logarithm?\n$log_{\frac{1}{2}}w^{4}u^{2}-v^{3}$\n$log_{\frac{1}{2}}w^{4}(\frac{u^{2}}{v^{3}})$\n$log_{\frac{1}{2}}(\frac{w^{4}}{u^{2}v^{3}})$\n$log_{\frac{1}{2}}(w(\frac{u^{2}}{v^{3}}))^{4}$

what is $4log_{\frac{1}{2}}w+(2log_{\frac{1}{2}}u - 3log_{\frac{1}{2}}v)$ written as a single logarithm?\n$log_{\frac{1}{2}}w^{4}u^{2}-v^{3}$\n$log_{\frac{1}{2}}w^{4}(\frac{u^{2}}{v^{3}})$\n$log_{\frac{1}{2}}(\frac{w^{4}}{u^{2}v^{3}})$\n$log_{\frac{1}{2}}(w(\frac{u^{2}}{v^{3}}))^{4}$

Answer

Explanation:

Step1: Apply power - rule of logarithms

The power - rule states that $n\log_aM=\log_aM^n$. For $4\log_{\frac{1}{2}}w$, we get $\log_{\frac{1}{2}}w^4$. For $2\log_{\frac{1}{2}}u$, we get $\log_{\frac{1}{2}}u^2$. For $3\log_{\frac{1}{2}}v$, we get $\log_{\frac{1}{2}}v^3$. So the expression becomes $\log_{\frac{1}{2}}w^4+(\log_{\frac{1}{2}}u^2 - \log_{\frac{1}{2}}v^3)$.

Step2: Use the sum and difference rules of logarithms

The sum rule is $\log_aM+\log_aN=\log_a(MN)$ and the difference rule is $\log_aM-\log_aN=\log_a\frac{M}{N}$. First, simplify the expression inside the parentheses: $\log_{\frac{1}{2}}u^2 - \log_{\frac{1}{2}}v^3=\log_{\frac{1}{2}}\frac{u^2}{v^3}$ (by the difference rule). Then, $\log_{\frac{1}{2}}w^4+\log_{\frac{1}{2}}\frac{u^2}{v^3}=\log_{\frac{1}{2}}(w^4\cdot\frac{u^2}{v^3})$ (by the sum rule).

Answer:

$\log_{\frac{1}{2}}w^4\left(\frac{u^2}{v^3}\right)$ (the second option)