which expression is equivalent to $log_{w}\frac{(x^{2}-6)^{4}}{sqrt3{x^{2}+8}}$?\n$4log_{w}\frac{x^{2}}{1296}…

which expression is equivalent to $log_{w}\frac{(x^{2}-6)^{4}}{sqrt3{x^{2}+8}}$?\n$4log_{w}\frac{x^{2}}{1296}-\frac{1}{3}log_{w}(2x + 8)$\n$4log_{w}(x^{2}-6)-3log_{w}(x^{2}+8)$\n$4log_{w}(x^{2}-6)-\frac{1}{3}log_{w}(x^{2}+8)$\n$4(log_{w}x^{2}-\frac{1}{3}log_{w}(x^{2}+8)-6)$

which expression is equivalent to $log_{w}\frac{(x^{2}-6)^{4}}{sqrt3{x^{2}+8}}$?\n$4log_{w}\frac{x^{2}}{1296}-\frac{1}{3}log_{w}(2x + 8)$\n$4log_{w}(x^{2}-6)-3log_{w}(x^{2}+8)$\n$4log_{w}(x^{2}-6)-\frac{1}{3}log_{w}(x^{2}+8)$\n$4(log_{w}x^{2}-\frac{1}{3}log_{w}(x^{2}+8)-6)$

Answer

Explanation:

Step1: Apply log - power rule

The power - rule of logarithms states that $\log_aM^n=n\log_aM$. For the numerator $\log_w(x^{2}-6)^4$, by the power - rule, we have $\log_w(x^{2}-6)^4 = 4\log_w(x^{2}-6)$.

Step2: Apply log - power rule to the denominator

For the denominator $\log_w\sqrt[3]{x^{2}+8}=\log_w(x^{2}+8)^{\frac{1}{3}}$. By the power - rule of logarithms, $\log_w(x^{2}+8)^{\frac{1}{3}}=\frac{1}{3}\log_w(x^{2}+8)$.

Step3: Use log - quotient rule

The quotient rule of logarithms states that $\log_a\frac{M}{N}=\log_aM-\log_aN$. So, $\log_w\frac{(x^{2}-6)^4}{\sqrt[3]{x^{2}+8}}=4\log_w(x^{2}-6)-\frac{1}{3}\log_w(x^{2}+8)$.

Answer:

$4\log_w(x^{2}-6)-\frac{1}{3}\log_w(x^{2}+8)$ (the third option)