# Math Help - Ugh... Logarithms =/

1. ## 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)

2. ## Re: Ugh... Logarithms =/

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:

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

dokrbb

3. ## Re: Ugh... Logarithms =/

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 \begin{align*} 3\ln{(x)} \left[ \ln{(x+3)} + \ln{(x-3)} \right] \end{align*}?

4. ## Re: Ugh... Logarithms =/

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 \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)}

5. ## Re: Ugh... Logarithms =/

\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*}