logaithm help.

**Iamnothere** · March 27th 2012, 12:57 PM

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

**Plato** · March 27th 2012, 01:09 PM

Originally Posted by Iamnothere

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

Because $\sqrt3>1$ we must rewwrite $\log_{25}(1-\sqrt3)^2=2\log_{25}(\sqrt3-1)$

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

**HallsofIvy** · March 27th 2012, 01:55 PM

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

**Iamnothere** · March 27th 2012, 01:56 PM

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

**HallsofIvy** · March 27th 2012, 01:59 PM

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 $\sqrt{3}> \sqrt{1}= 1$ . In fact, it is not at all difficult to use a calculator to see that $\sqrt{3}= 1.7320508075688772935274463415059... > 1$ .

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

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

**Iamnothere** · March 27th 2012, 02:00 PM

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.

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