Just want to make sure about this logarithm problem

**dc52789** · October 25th 2008, 09:47 AM

I had a problem similar to this one.

Write expression as a logarithm of a single quantity and then simplify if possible.

(1/4)[log (x²-1)-log (x+1)]+3log x

Please show the steps.

**11rdc11** · October 25th 2008, 10:20 AM

Originally Posted by dc52789

I had a problem similar to this one.

Write expression as a logarithm of a single quantity and then simplify if possible.

(1/4)[log (x²-1)-log (x+1)]+3log x

Please show the steps.

Use log properties

$\log{x}^y=y\log{x}$

$\log{xy}=\log{x}+\log{y}$

$\log{\frac{x}{y}}=\log{x}-\log{y}$

**dc52789** · October 25th 2008, 10:24 AM

Originally Posted by 11rdc11

it simplfies to $\log{(x^4-x^3)}$

Um...I still need to know the steps to how you get that answer.

**Moo** · October 25th 2008, 10:25 AM

Hello,

Originally Posted by dc52789

I had a problem similar to this one.

Write expression as a logarithm of a single quantity and then simplify if possible.

(1/4)[log (x²-1)-log (x+1)]+3log x

Please show the steps.

Note that x>0 and |x|>1, so x>1. Otherwise it is not defined.

notice that x²-1=(x-1)(x+1)

So $\log(x^2-1)-\log(x+1)=\log \left(\frac{x^2-1}{x+1}\right)=\log(x-1)$

$\frac 14 \cdot \log(x-1)+3 \log(x)=\log \left((x-1)^{1/4}\right)+\log(x^3)$

$=\log \left((x-1)^{1/4} \cdot x^3\right)=\log \left((x^{13}-x^{12})^{1/4}\right)$

now it depends on how you want it to be simplified.

**dc52789** · October 25th 2008, 10:37 AM

Originally Posted by Moo

Hello,

Note that x>0 and |x|>1, so x>1. Otherwise it is not defined.

notice that x²-1=(x-1)(x+1)

So $\log(x^2-1)-\log(x+1)=\log \left(\frac{x^2-1}{x+1}\right)=\log(x-1)$

$\frac 14 \cdot \log(x-1)+3 \log(x)=\log \left((x-1)^{1/4}\right)+\log(x^3)$

$=\log \left((x-1)^{1/4} \cdot x^3\right)=\log \left((x^{13}-x^{12})^{1/4}\right)$

now it depends on how you want it to be simplified.

o k...which would be an easier way to simplify it?

**dc52789** · October 25th 2008, 10:58 AM

Originally Posted by 11rdc11

$=\log \left((x-1)^{1/4} \cdot x^3\right)=\log \left((x^{13}-x^{12})^{1/4}\right)$

Moo how are these equal?

I don't know. I get lost whenever I have these questions.

**11rdc11** · October 25th 2008, 11:09 AM

Originally Posted by Moo

Hello,

Note that x>0 and |x|>1, so x>1. Otherwise it is not defined.

notice that x²-1=(x-1)(x+1)

So $\log(x^2-1)-\log(x+1)=\log \left(\frac{x^2-1}{x+1}\right)=\log(x-1)$

$\frac 14 \cdot \log(x-1)+3 \log(x)=\log \left((x-1)^{1/4}\right)+\log(x^3)$

$=\log \left((x-1)^{1/4} \cdot x^3\right)=\log \left((x^{13}-x^{12})^{1/4}\right)$

now it depends on how you want it to be simplified.

$=\log \left((x-1)^{1/4} \cdot x^3\right)=\log \left((x^{13}-x^{12})^{1/4}\right)$

Here is how moo did it

$\log \left((x^{13}-x^{12})^{1/4}\right)$

$~^4\sqrt{x^{12}(x-1)}$

$x^3(^4\sqrt{x-1})$

**Moo** · October 25th 2008, 11:12 AM

Originally Posted by 11rdc11

$=\log \left((x-1)^{1/4} \cdot x^3\right)=\log \left((x^{13}-x^{12})^{1/4}\right)$

Here is how moo did it

$\log \left((x^{13}-x^{12})^{1/4}\right)$

$~^4\sqrt{x^{12}(x-1)}$

$x^3(\sqrt^4{x-1})$

Not exactly the way I did^^

$x^3=(x^{12})^{1/4}$ and then you can distribute it in the $(x-1)^{1/4}$ part

By the way, dc52789, it would be nice if you thought a little instead of just asking all the steps =)

**dc52789** · October 25th 2008, 11:18 AM

Originally Posted by Moo

Not exactly the way I did^^

$x^3=(x^{12})^{1/4}$ and then you can distribute it in the $(x-1)^{1/4}$ part

By the way, dc52789, it would be nice if you thought a little instead of just asking all the steps =)

I know but it really got me lost and the professor didn't go over it before the test.

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