• Oct 25th 2008, 10:47 AM
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

• Oct 25th 2008, 11:20 AM
11rdc11
Quote:

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

Use log properties

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

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

$\log{\frac{x}{y}}=\log{x}-\log{y}$
• Oct 25th 2008, 11:24 AM
dc52789
Quote:

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.
• Oct 25th 2008, 11:25 AM
Moo
Hello,
Quote:

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

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.
• Oct 25th 2008, 11:37 AM
dc52789
Quote:

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?
• Oct 25th 2008, 11:58 AM
dc52789
Quote:

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.
• Oct 25th 2008, 12:09 PM
11rdc11
Quote:

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

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

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

$x^3(^4\sqrt{x-1})$
• Oct 25th 2008, 12:12 PM
Moo
Quote:

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 =)
• Oct 25th 2008, 12:18 PM
dc52789
Quote:

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.