which expression is equivalent to $log_{12}\frac{x^{4}sqrt{x^{3}-2}}{(x + 1)^{5}}$?\n$4log_{12}x+\frac{1}{2}l…

which expression is equivalent to $log_{12}\frac{x^{4}sqrt{x^{3}-2}}{(x + 1)^{5}}$?\n$4log_{12}x+\frac{1}{2}log_{12}(x^{3}-2)-5log_{12}(x + 1)$\n$4log_{12}x+\frac{1}{2}log_{12}\frac{x^{3}}{2}-5log_{12}x+log_{12}1$\n$log_{12}4x+\frac{1}{2}log_{12}(x^{3}-2)-5log_{12}(x)+1$\n$4log_{12}x+\frac{1}{2}log_{12}(x^{3}-2)-5log_{12}(x + 1)$

which expression is equivalent to $log_{12}\frac{x^{4}sqrt{x^{3}-2}}{(x + 1)^{5}}$?\n$4log_{12}x+\frac{1}{2}log_{12}(x^{3}-2)-5log_{12}(x + 1)$\n$4log_{12}x+\frac{1}{2}log_{12}\frac{x^{3}}{2}-5log_{12}x+log_{12}1$\n$log_{12}4x+\frac{1}{2}log_{12}(x^{3}-2)-5log_{12}(x)+1$\n$4log_{12}x+\frac{1}{2}log_{12}(x^{3}-2)-5log_{12}(x + 1)$

Answer

Explanation:

Step1: Apply log - quotient rule

The quotient rule of logarithms is $\log_a\frac{M}{N}=\log_aM-\log_aN$. So, $\log_{12}\frac{x^{4}\sqrt{x^{3}-2}}{(x + 1)^{5}}=\log_{12}(x^{4}\sqrt{x^{3}-2})-\log_{12}(x + 1)^{5}$.

Step2: Apply log - product rule

The product rule of logarithms is $\log_a(MN)=\log_aM+\log_aN$. Then $\log_{12}(x^{4}\sqrt{x^{3}-2})=\log_{12}x^{4}+\log_{12}\sqrt{x^{3}-2}$.

Step3: Rewrite square - root as exponent

Since $\sqrt{x^{3}-2}=(x^{3}-2)^{\frac{1}{2}}$, we have $\log_{12}\sqrt{x^{3}-2}=\log_{12}(x^{3}-2)^{\frac{1}{2}}$.

Step4: Apply log - power rule

The power rule of logarithms is $\log_aM^{n}=n\log_aM$. So, $\log_{12}x^{4}=4\log_{12}x$, $\log_{12}(x^{3}-2)^{\frac{1}{2}}=\frac{1}{2}\log_{12}(x^{3}-2)$ and $\log_{12}(x + 1)^{5}=5\log_{12}(x + 1)$. Combining these results, we get $\log_{12}\frac{x^{4}\sqrt{x^{3}-2}}{(x + 1)^{5}}=4\log_{12}x+\frac{1}{2}\log_{12}(x^{3}-2)-5\log_{12}(x + 1)$.

Answer:

$4\log_{12}x+\frac{1}{2}\log_{12}(x^{3}-2)-5\log_{12}(x + 1)$ (the fourth option)