# Ugh... Logarithms =/

• Apr 21st 2013, 05:09 PM
swishh
Ugh... Logarithms =/
Hello y'all, I hope y'all are having an excellent weekend.

Can you help with this logarithm nproblem?

Condense the expression to the logarithm of a single quantity

3lnx- (ln(x+3) + ln(x-3)
• Apr 21st 2013, 05:32 PM
dokrbb
Re: Ugh... Logarithms =/
Quote:

Originally Posted by swishh
Hello y'all, I hope y'all are having an excellent weekend.

Can you help with this logarithm nproblem?

Condense the expression to the logarithm of a single quantity

3lnx- (ln(x+3) + ln(x-3)

you would be willing to use these two identities:

$\displaystyle \log_b(xy) = \log_b(x) + \log_b(y)$ and this $\displaystyle \log_b\frac{x}{y} = \log_b(x) - \log_b(y)$,

dokrbb
• Apr 22nd 2013, 01:04 AM
Prove It
Re: Ugh... Logarithms =/
Quote:

Originally Posted by swishh
Hello y'all, I hope y'all are having an excellent weekend.

Can you help with this logarithm nproblem?

Condense the expression to the logarithm of a single quantity

3lnx- (ln(x+3) + ln(x-3)

It helps to check that you have closed all sets of brackets. Just so we're clear, is your expression \displaystyle \displaystyle \begin{align*} 3\ln{(x)} \left[ \ln{(x+3)} + \ln{(x-3)} \right] \end{align*}?
• Apr 22nd 2013, 04:25 PM
swishh
Re: Ugh... Logarithms =/
Quote:

Originally Posted by Prove It
It helps to check that you have closed all sets of brackets. Just so we're clear, is your expression \displaystyle \displaystyle \begin{align*} 3\ln{(x)} \left[ \ln{(x+3)} + \ln{(x-3)} \right] \end{align*}?

sorry for the late reply! i still need help, I don't have much longer! I'll rewrite the problem, there was a MISTAKE on my end in the OP, i'm not sure if you have it right. Thank you in advance to anyone who can briefly go over the steps.

2lnx - {ln(x+2) + ln(x-2)}
• Apr 22nd 2013, 04:50 PM
Prove It
Re: Ugh... Logarithms =/
\displaystyle \displaystyle \begin{align*} 2\ln{(x)} - \left[ \ln{(x + 2)} + \ln{(x - 2)} \right] &= \ln{ \left( x^2 \right) } - \ln{\left[ (x+2)(x-2) \right]} \\ &= \ln{\left[ \frac{x^2}{(x+2)(x-2)} \right] } \end{align*}