Question

Given ${\log }_b\left(a\right)=x$ and ${\log }_b\left(c\right)=y,{\log }_{a^2}\left(\sqrt[3]{3b^5{c}^4}\right)=$

• A. $\frac{\frac{5}{3}+y^4}{x^2}$
• B. $\frac{5+4y}{6x}$
• C. $\frac{20y}{3x}$
• D. $2x+4y$
• E. $2x+\frac{20y}{3}$

Answer 1

The correct option is B $\frac{5+4y}{6x}$

It is given that ${\log }_b\left(a\right)=x$ and ${\log }_b\left(c\right)=y$.

Use the change of base formula to rewrite ${\log }_{a^2}\left(\sqrt[3]{3b^5{c}^4}\right)$ as:

${\log }_{a^2}\left(\sqrt[3]{3b^5{c}^4}\right)=\frac{{\log }_b\left(\sqrt[3]{3b^5{c}^4}\right)}{{\log }_b\left(a^2\right)}$

Apply the properties of logarithms to simplify the numerator and denominator of the fraction:

$\frac{{\log }_b\left(\sqrt[3]{3b^5{c}^4}\right)}{{\log }_b\left(a^2\right)}=\frac{{\log }_b(b^5{c}^4{)}^{\frac{1}{3}}}{{\log }_b\left(a^2\right)}$

$=\frac{\frac{1}{3}{\log }_b\left(b^5{c}^4\right)}{2{\log }_b\left(a\right)}$

$=\frac{\frac{1}{3}{\log }_b\left(b^5\right)+\frac{1}{3}{\log }_b\left(c^4\right)}{2{\log }_b\left(a\right)}$

$=\frac{\frac{5}{3}{\log }_b\left(b\right)+\frac{4}{3}{\log }_b\left(c\right)}{2{\log }_b\left(a\right)}$

$=\frac{\frac{5}{3}+\frac{4}{3}y}{2x}$

$=\frac{5+4y}{6x}$