# Thread: Expanding logarithms

1. ## Expanding logarithms

I understand the laws of logarithms, they are very simple and I also know how to expand logs but this one has got me stumped because there are 3 square roots. So here is my question.

Use the Laws of Logarithms to expand the expression.

I know I can raise the equation to the $\displaystyle 1/2$ power to get rid of the first square root but then I'm stuck.

2. $\displaystyle \ln\left(\sqrt{x^4\sqrt{y\sqrt{z^2}}}\right) =$

$\displaystyle \frac{1}{2}\ln\left(x^4\sqrt{y\sqrt{z^2}}\right) =$

$\displaystyle \frac{1}{2}\ln(x^4) + \frac{1}{2} \ln\left(\sqrt{y\sqrt{z^2}}\right) =$

$\displaystyle \frac{1}{2}\ln(x^4) + \frac{1}{4} \ln\left(y\sqrt{z^2}\right) =$

$\displaystyle \frac{1}{2}\ln(x^4) + \frac{1}{4} \ln(y) + \frac{1}{4}\ln\left(\sqrt{z^2}\right)$

finish up?

3. Yeah I should be able to finish it up. I'm just wondering on the 3'rd step why you put $\displaystyle 1/2$ in front of 2'nd part of the equation. Then for the 4th step you just multiplied the $\displaystyle 1/2$ by the $\displaystyle 1/2$ that the second part of the equation would've been raised to?

4. step 3, justified ...

$\displaystyle \frac{1}{2}\ln\left(x^4\sqrt{y\sqrt{z}}\right) =$

$\displaystyle \frac{1}{2}\left[\ln\left(x^4\sqrt{y\sqrt{z^2}}\right)\right] =$

$\displaystyle \frac{1}{2}\left[\ln(x^4) + \ln\left(\sqrt{y\sqrt{z^2}}\right)\right] =$

$\displaystyle \frac{1}{2}\ln(x^4) + \frac{1}{2}\ln\left(\sqrt{y\sqrt{z^2}}\right)$

for the 4th step, the outer radical only affects the second log term, not the first.

5. All right I get it, thanks. I appreciate explaining it to me. The answer is: