## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 4 - Section 4.4 - Laws of Logarithms - 4.4 exercises: 48

#### Answer

$\log\sqrt{x\sqrt{y\sqrt{z}}}=\dfrac{1}{2}\log x+\dfrac{1}{4}\log y+\dfrac{1}{8}\log z$

#### Work Step by Step

$\log\sqrt{x\sqrt{y\sqrt{z}}}$ Let's rewrite the expression like this: $\log(x\sqrt{y\sqrt{z}})^{1/2}$ Let's take the exponent to the front of the logarithm to multiply: $\log(x\sqrt{y\sqrt{z}})^{1/2}=\dfrac{1}{2}\log(x\sqrt{y\sqrt{z}})=...$ The logarithm of a product can be expanded as a sum: $...=\dfrac{1}{2}[\log x+\log\sqrt{y\sqrt{z}}]=...$ Let's rewrite the expression: $...=\dfrac{1}{2}[\log x+\log(y\sqrt{z})^{1/2}]=...$ We can take the exponent in $\log(y\sqrt{z})^{1/2}$ to multiply to the front of the logarithm: $...=\dfrac{1}{2}[\log x+\dfrac{1}{2}\log(y\sqrt{z})]=...$ Again, we expand the logarithm of the product as a sum: $...=\dfrac{1}{2}[\log x+\dfrac{1}{2}(\log y+\log\sqrt{z})]=...$ Rewrite the expression like this: $...=\dfrac{1}{2}[\log x+\dfrac{1}{2}(\log y+\log z^{1/2})]=...$ Finally, take the exponent in $\log z^{1/2}$ to multiply to the front of the logarithm: $...=\dfrac{1}{2}[\log x+\dfrac{1}{2}(\log y+\dfrac{1}{2}\log z)]=...$ $...=\dfrac{1}{2}\log x+\dfrac{1}{4}\log y+\dfrac{1}{8}\log z$

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