How do I simplify this? $\displaystyle ln(x^9 \sqrt{4-x^2})$ Thanks

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- Feb 18th 2008, 05:05 PMzsig013Simplify Natural Logarithm
How do I simplify this? $\displaystyle ln(x^9 \sqrt{4-x^2})$ Thanks

- Feb 18th 2008, 05:10 PMtopsquark
You know that $\displaystyle ln(ab) = ln(a) + ln(b)$ so

$\displaystyle ln(x^9 \sqrt{4-x^2}) = ln(x^9) + ln(\sqrt{4 - x^2})$

Rewrite that last term:

$\displaystyle = ln(x^9) + ln((4 - x^2)^{1/2})$

Now recall that $\displaystyle ln(a^b) = b \cdot ln(a)$, sooooooo.....

$\displaystyle = 9 \cdot ln(x) + \frac{1}{2} \cdot ln(4 - x^2)$

Now, you could finish here, but there's one last thing we can do:

$\displaystyle 4 - x^2 = (2 + x)(2 - x)$, so looking at that last term again:

$\displaystyle = 9 \cdot ln(x) + \frac{1}{2} \cdot ln((2 + x)(2 - x))$

$\displaystyle = 9 \cdot ln(x) + \frac{1}{2} \cdot ln(2 + x) + \frac{1}{2} \cdot ln(2 - x)$

-Dan - Feb 18th 2008, 05:12 PMzsig013
Thanks a lot.. I was thinking along the same lines there, but wouldn't that be considered expanded?

- Feb 18th 2008, 05:23 PMtopsquark
The term "simplify" is kind of hard to define. The result of simplifying an expression depends heavily on what use you have for the form of the final answer. In this case I am assuming that we are going to want the arguments of the ln function to have the lowest degree possible.

-Dan - Feb 18th 2008, 05:28 PMzsig013
Glad you cleared that up..(Clapping) I submitted my quiz and got it right. I was bashing my head on the keyboard for half an hour trying to figure out how to "simplify" that problem. I initially had that written down, but thought nah that can't be it.(Headbang)