# Logarithm question

• Apr 6th 2010, 09:19 AM
bolsius
Logarithm question

• Apr 6th 2010, 10:33 AM
CaptainBlack
Quote:

Originally Posted by bolsius

CB
• Apr 6th 2010, 10:33 AM
Quote:

Originally Posted by bolsius

lol.

Can't see anything.

But while I'm here I may as well throw down some properties of logs in case they help...

$\displaystyle \log (ab) = \log(a) + \log(b)$.

$\displaystyle \log(a/b) = \log(a) - \log(b)$ as long as $\displaystyle b \neq 0$.

$\displaystyle \log(e^x) = x$ where [tex]e[/math[ is the exponential function.
• Apr 6th 2010, 02:12 PM
bolsius
Quote:

lol.

Can't see anything.

But while I'm here I may as well throw down some properties of logs in case they help...

$\displaystyle \log (ab) = \log(a) + \log(b)$.

$\displaystyle \log(a/b) = \log(a) - \log(b)$ as long as $\displaystyle b \neq 0$.

$\displaystyle \log(e^x) = x$ where [tex]e[/math[ is the exponential function.

All right sorry it was a copy/paste and I dont know why it didnīt show up so I write it here

ln (1-x) + ln (1+x) = ln (3/4)

thanks
• Apr 6th 2010, 02:18 PM
CaptainBlack
Quote:

Originally Posted by bolsius
All right sorry it was a copy/paste and I dont know why it didnīt show up so I write it here

ln (1-x) + ln (1+x) = ln (3/4)

thanks

Use the laws of logarithms to combine the two logs on the left:

$\displaystyle \ln(1-x)+\ln(1+x)=\ln((1-x)(1+x))=\ln(1-x^2)=\ln(3/4)$

can you take it from there?

CB
• Apr 6th 2010, 02:19 PM
Quote:

Originally Posted by bolsius
All right sorry it was a copy/paste and I dont know why it didnīt show up so I write it here

ln (1-x) + ln (1+x) = ln (3/4)

thanks

$\displaystyle \ln(a) + \ln(b) = \ln(ab)$ (remember this rule).

So you have...

$\displaystyle \ln(1-x) + \ln(1+x) = \ln((1-x)(1+x)) = \ln(1 - x^2) = \ln(3/4)$

Solve for x.

EDIT: too slow...
• Apr 6th 2010, 02:43 PM
bolsius
Yes thanks I got it...I wasnīt sure if I had to use the laws of logarithms both side of the equal sign (right and left)