Question

Express the given logarithm as a sum and/or difference of logarithms. Simplify, if possible. assume that all variables represent positive real numbers.

Answer 1

A

`(1)/(8)\log _9r+(1)/(5)\log _9s-2\log _9u^{}`

Step-by-step explanation

Step 1

remember some properties of the logarithms:

`\begin{gathered} \log (A\cdot B)=\log \text{ A+log B} \\ \log ((a)/(b))=\log \text{ A- log B} \\ \log a^b=b\cdot\text{ log a} \\ \log \sqrt[b]{a}\text{ = }\frac{\text{log A}}{b} \end{gathered}`

then

`\begin{gathered} \log _9\frac{\sqrt[8]{r}\sqrt[5]{s}}{u^2} \\ \log _9\frac{\sqrt[8]{r}\sqrt[5]{s}}{u^2}=\log _9\sqrt[8]{r}\sqrt[5]{s}-log_9u^2 \\ \log _9\frac{\sqrt[8]{r}\sqrt[5]{s}}{u^2}=\log _9\sqrt[8]{r}+\log _9\sqrt[5]{s}-log_9u^2 \\ \log _9\frac{\sqrt[8]{r}\sqrt[5]{s}}{u^2}=\log _9r^{(1)/(8)}+\log _9s^{(1)/(5)}-2\log _9u \\ \log _9\frac{\sqrt[8]{r}\sqrt[5]{s}}{u^2}=(1)/(8)\log _9r+(1)/(5)\log _9s-2\log _9u^{} \end{gathered}`

I hope this helps you