# Rewriting expression as single log with coefficent 1

• Jul 7th 2009, 02:05 PM
Snowcrash
Rewriting expression as single log with coefficent 1
I am rewriting the following as a single log with a coefficient of 1:

ln((x^3)-1) - ln((x^2) + x + 1)

Because the coefficient is already zero, I believe I need to divide the two expressions:

ln ( ((x^3)-1) / ((x^2) + x + 1) )

And cancel out the ones...

ln ( (x^3) / ((x^2) + x) )

This seems too easy - can I break it down further?
• Jul 7th 2009, 02:14 PM
Plato
Quote:

Originally Posted by Snowcrash
I am rewriting the following as a single log with a coefficient of 1:

ln((x^3)-1) - ln((x^2) + x + 1)

Because the coefficient is already zero, I believe I need to divide the two expressions:

ln ( ((x^3)-1) / ((x^2) + x + 1) )

And cancel out the ones...

ln ( (x^3) / ((x^2) + x) )

You know that cannot be done.
Only factors can be divided out.
But this is true $\displaystyle (x^3-1)=(x-1)(x^2+x+1)$

So this is also true $\displaystyle \ln \left( {\frac{{x^3 - 1}}{{x^2 + x + 1}}} \right) = \ln \left( {x - 1} \right)$. WHY?
• Jul 7th 2009, 02:15 PM
red_dog
$\displaystyle \ln\frac{x^3-1}{x^2+x+1}=\ln\frac{(x-1)(x^2+x+1)}{x^2+x+1}=\ln(x-1)$
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