1. ## logarithm help.

How do you solve this? Why does the answer become squareroot(3)-1??

2. ## Re: logaithm help.

Originally Posted by Iamnothere
How do you solve this? Why does the answer become squareroot(3)-1??
Because $\displaystyle \sqrt3>1$ we must rewwrite $\displaystyle \log_{25}(1-\sqrt3)^2=2\log_{25}(\sqrt3-1)$

Note $\displaystyle 25^{\log_{25}(\sqrt3-1)}=\sqrt3 -1$

3. ## Re: logaithm help.

$\displaystyle 5= 25^{1/2}$ so
$\displaystyle 5^{log_{25}(a)}= (25^{1/2})^{log_{25}(a)}= 25^{(1/2)log_{25}(a)}= 25^{log_{25}(\sqrt{a})}= \sqrt{a}$

4. ## Re: logaithm help.

Do you mind me asking where does that rule come from? The one with sqrt(3)>1, I'm a little bit confused.

5. ## Re: logaithm help.

Originally Posted by Iamnothere
Do you mind me asking where does that rule come from? The one with sqrt(3)>1, I'm a little bit confused.
??? You will accept 3> 1 won't you? And square root is an increasing function so $\displaystyle \sqrt{3}> \sqrt{1}= 1$. In fact, it is not at all difficult to use a calculator to see that $\displaystyle \sqrt{3}= 1.7320508075688772935274463415059... > 1$.

Or do you mean "$\displaystyle ln(1- \sqrt{3})^2)= 2ln(\sqrt{3}- 1)$"?

Obviously both $\displaystyle (1- \sqrt{3})^2$ and $\displaystyle (\sqrt{3}- 1)^2$ give the same thing and $\displaystyle log(a^2)= 2log(a)$ provided that a is positive becauser log(a) is not defined for negative a.

6. ## Re: logaithm help.

Emm, I think you got me a little bit wrong here, I was just wondering why did we have to write sqrt(3)-1 instead of 1-sqrt(3).

Edit: Ah yeah, I totally forgot about looking at that part. Thank you.