Write the expression as the logarithm of a single quantity

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February 15th 2009, 05:21 PM
x0ne

Write the expression as the logarithm of a single quantity

Hello,

The problem is this:
2[ln(x) - ln(x + 1) - ln(x - 1)]

I know to distribute the 2 first:
2ln(x) - 2ln(x+1) - 2ln(x-1)

Then take the 2 and make it an exponent:
ln(x)^2 - ln(x+1)^2 - ln(x-1)^2

I know at this point since subtraction is being used we will divide each so:

ln[ x^2 / (x+1)^2] - ln(x-1)^2

I am lost right there and not sure what to do next. Any help would be greatly appreciated.
February 15th 2009, 05:31 PM
mollymcf2009

Quote:

Originally Posted by x0ne

Hello,

The problem is this:
2[ln(x) - ln(x + 1) - ln(x - 1)]

I know to distribute the 2 first:
2ln(x) - 2ln(x+1) - 2ln(x-1)

Then take the 2 and make it an exponent:
ln(x)^2 - ln(x+1)^2 - ln(x-1)^2

I know at this point since subtraction is being used we will divide each so:

ln[ x^2 / (x+1)^2] - ln(x-1)^2

I am lost right there and not sure what to do next. Any help would be greatly appreciated.

When you have a fraction divided by a fraction, you can multiply the top by the reciprocal of the bottom.

$\frac{\frac{x}{x+1}}{x-1}$

$\frac{x}{x+1} \cdot \frac{1}{x-1}$

Can you get it from there?
February 15th 2009, 05:37 PM
x0ne

Quote:

Originally Posted by mollymcf2009

When you have a fraction divided by a fraction, you can multiply the top by the reciprocal of the bottom.

$\frac{\frac{x}{x+1}}{x-1}$

$\frac{x}{x+1} \cdot \frac{1}{x-1}$

Can you get it from there?

Alright. So in my case my problem would now turn into:

$\frac{x^2}{(x+1) (x+1)} \cdot \frac{1}{(x-1) (x-1)}$

Would that be the proper format?
February 15th 2009, 05:46 PM
mollymcf2009

Quote:

Originally Posted by x0ne

Alright. So in my case my problem would now turn into:

$\frac{x^2}{(x+1) (x+1)} \cdot \frac{1}{(x-1) (x-1)}$

Would that be the proper format?

No, you are just multiplying the top by the reciprocal of the bottom to get rid of your big fraction. It's just to simplify it.

You divided your first two logs correctly, then you needed to divide that by the last subtracted log, from your logs rules, right?

$\frac{x}{x+1} \cdot \frac{1}{x-1}$
$<br /> = \frac{x}{x^2-1}$

So your log will look like this:

$ln(\frac{x}{x^2-1})^2$

Does that make sense to you?
February 15th 2009, 05:52 PM
x0ne

Quote:

Originally Posted by mollymcf2009

No, you are just multiplying the top by the reciprocal of the bottom to get rid of your big fraction. It's just to simplify it.

You divided your first two logs correctly, then you needed to divide that by the last subtracted log, from your logs rules, right?

$\frac{x}{x+1} \cdot \frac{1}{x-1}$
$<br /> = \frac{x}{x^2-1}$

So your log will look like this:

$ln(\frac{x}{x^2-1})^2$

Does that make sense to you?

Yep. I see what you did. Because everything was being squared there was no need to represent it twice. Instead use one of the values, solve the equation and put the exponent back.