Question

score: 13.67/28 answered: 6/10 question 6 expand each of the following logarithms using the product rule, quotient rule, and/or power rule. do not use the change of base formula. \\(\ln \left(8y^{2}a^{3}\
ight) = \square\\) \\(\log_{4} \left(vw^{2}\
ight)^{6} = \square\\) question help: video submit question

Explanation:

Step1: Expand $\ln(8y^2a^3)$

Use the product rule of logarithms: $\ln(xyz)=\ln x+\ln y+\ln z$. So, $\ln(8y^2a^3)=\ln 8+\ln y^2+\ln a^3$. Then, use the power rule: $\ln x^n = n\ln x$. So, $\ln 8+\ln y^2+\ln a^3=\ln 8 + 2\ln y+3\ln a$. (Note: $\ln 8$ can also be written as $3\ln 2$ since $8 = 2^3$, so it can be further expanded as $3\ln 2+2\ln y + 3\ln a$)

Step2: Expand $\log_4(vw^2)^6$

First, use the power rule of logarithms: $\log_b x^n=n\log_b x$. So, $\log_4(vw^2)^6 = 6\log_4(vw^2)$. Then, use the product rule: $\log_b(xy)=\log_b x+\log_b y$. So, $6\log_4(vw^2)=6(\log_4 v+\log_4 w^2)$. Again, use the power rule: $6(\log_4 v+\log_4 w^2)=6\log_4 v + 6\times2\log_4 w=6\log_4 v+12\log_4 w$

Answer:

For $\ln(8y^2a^3)$: $\boldsymbol{\ln 8 + 2\ln y+3\ln a}$ (or $\boldsymbol{3\ln 2 + 2\ln y+3\ln a}$)
For $\log_4(vw^2)^6$: $\boldsymbol{6\log_4 v + 12\log_4 w}$ (or intermediate steps like $6\log_4 v+6\log_4 w^2$ or $6(\log_4 v+\log_4 w^2)$ are also correct based on the level of expansion, but the fully expanded with power rule applied to $w^2$ is $6\log_4 v + 12\log_4 w$)