Logarithm question

**bolsius** · April 6th 2010, 09:19 AM

Can someone please help me start on this:

thanks

**CaptainBlack** · April 6th 2010, 10:33 AM

Originally Posted by bolsius

Can someone please help me start on this:

thanks

Broken link

CB

**Deadstar** · April 6th 2010, 10:33 AM

Originally Posted by bolsius

Can someone please help me start on this:

thanks

lol.

Can't see anything.

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

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

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

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

**bolsius** · April 6th 2010, 02:12 PM

Originally Posted by Deadstar

lol.

Can't see anything.

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

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

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

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

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

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

thanks

**CaptainBlack** · April 6th 2010, 02:18 PM

Originally Posted by bolsius

All right sorry it was a copy/paste and I dont know why it didn´t 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:

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

can you take it from there?

CB

**Deadstar** · April 6th 2010, 02:19 PM

Originally Posted by bolsius

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

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

thanks

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

So you have...

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

Solve for x.

EDIT: too slow...

**bolsius** · April 6th 2010, 02:43 PM

Yes thanks I got it...I wasn´t sure if I had to use the laws of logarithms both side of the equal sign (right and left)

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