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-   -   Help Evaluating Logarithm (http://www.physicsforums.com/showthread.php?t=580305)

theintarnets Feb22-12 03:33 PM

Help Evaluating Logarithm
 
1. The problem statement, all variables and given/known data
Evaluate without a calculator:
(log34 + log29)2 - (log34 - log29)2


2. Relevant equations



3. The attempt at a solution
(log34 + log29)2 - (log34 - log29)2

(2log32 + 2log23)2 - (2log32 - 2log23)2

And now I'm stuck....

rock.freak667 Feb22-12 04:39 PM

Re: Help Evaluating Logarithm
 
Quote:

Quote by theintarnets (Post 3778489)
3. The attempt at a solution
(log34 + log29)2 / (log34 - log29)2

(2log34 + 2log29) / (2log34 - 2log29)

And now I'm stuck....

You went from a minus to a divide, I think you confused it with log(a/b)= loga-logb

Try using this fact a2-b2 = (a+b)(a-b).

Karamata Feb22-12 04:40 PM

Re: Help Evaluating Logarithm
 
[tex]a^2-b^2=(a-b)(a+b)[/tex]


EDIT: how to delete message

eumyang Feb22-12 04:46 PM

Re: Help Evaluating Logarithm
 
Quote:

Quote by theintarnets (Post 3778489)
(log34 + log29)2 / (log34 - log29)2

(2log34 + 2log29) / (2log34 - 2log29)

It also looks like you went from
(log34 + log29)2
to
(log342 + log292),
which is not true.

theintarnets Feb22-12 04:50 PM

Re: Help Evaluating Logarithm
 
Ohhhhh, I see. I shall re-attempt now. Thanks.

Joffan Feb22-12 04:52 PM

Re: Help Evaluating Logarithm
 
Hmmm. I think it's simpler than that, guys -
Setting A = log34 and B = log29

(A+B)2 - (A-B)2 = (A2 + 2AB + B2) - (A2 - 2AB + B2)

...

and later on using that lognxk = k.lognx
and that logab [itex]\times[/itex] logbc = logac

theintarnets Feb22-12 04:52 PM

Re: Help Evaluating Logarithm
 
Erm...Nevermind. I'm still stuck. Can someone walk me through it please?

Edit: I totally forgot about factoring. I'll try that, thanks!!

jedishrfu Feb22-12 04:59 PM

Re: Help Evaluating Logarithm
 
try this (A+B)^2 - (A-B)^2 = X^2 - Y^2 = ( X + Y ) ( X - Y )

where X=(A+B) and Y=(A-B)

and you get (A+B)^2 - (A-B)^2 =(2A) (2B) = 4AB

theintarnets Feb22-12 05:02 PM

Re: Help Evaluating Logarithm
 
Okay one of my friends just told me to try changing the base. I did that, and now I have a giant mess on my hands. I have

(ln2*ln4 + ln3*ln9 / ln3*ln2)^2 - (ln2*ln4 - ln3*ln9 / ln3*ln2)^2 and somehow I'm supposed to get 16 from all of that. I'm not really sure how...

eumyang Feb22-12 05:04 PM

Re: Help Evaluating Logarithm
 
Quote:

Quote by theintarnets (Post 3778678)
Okay one of my friends just told me to try changing the base. I did that, and now I have a giant mess on my hands. I have

(ln2*ln4 + ln3*ln9 / ln3*ln2)^2 - (ln2*ln4 - ln3*ln9 / ln3*ln2)^2 and somehow I'm supposed to get 16 from all of that. I'm not really sure how...

No, no, NO! Don't change the base at the beginning. Do what Joffan suggested. You can change the base much later if you really want to.

theintarnets Feb22-12 05:08 PM

Re: Help Evaluating Logarithm
 
Okie. Trying again now.

Edit: I just can't seem to do it no matter what I try :(

eumyang Feb22-12 05:24 PM

Re: Help Evaluating Logarithm
 
Show us what you have so far.

theintarnets Feb22-12 05:46 PM

Re: Help Evaluating Logarithm
 
Well I tried applying this:
Setting A = log34 and B = log29
(A+B)2 - (A-B)2 = (A2 + 2AB + B2) - (A2 - 2AB + B2)
The A2's and B2's will cancel out leaving me with 4AB which would be

4(log34 * log29)

The answer in the book says I'm supposed to get 16. I tried changing the base to get
4(ln4*ln9 / ln3*ln2)
I think I could maybe do 4(2ln2*2ln3 / ln3*ln2) but I'm not sure if that's correct. I think then maybe the ln2's and ln3's would cancel out leaving me with 4(2*2) which would be 16. But I'm not sure if that's correct.

eumyang Feb22-12 05:56 PM

Re: Help Evaluating Logarithm
 
Quote:

Quote by theintarnets (Post 3778756)
Well I tried applying this:
Setting A = log34 and B = log29
(A+B)2 - (A-B)2 = (A2 + 2AB + B2) - (A2 - 2AB + B2)
The A2's and B2's will cancel out leaving me with 4AB which would be

4(log34 * log29)

The answer in the book says I'm supposed to get 16. I tried changing the base to get
4(ln4*ln9 / ln3*ln2)
I think I could maybe do 4(2ln2*2ln3 / ln3*ln2) but I'm not sure if that's correct. I think then maybe the ln2's and ln3's would cancel out leaving me with 4(2*2) which would be 16. But I'm not sure if that's correct.

That's correct.

theintarnets Feb22-12 06:39 PM

Re: Help Evaluating Logarithm
 
Yay!!! Thanks so much everyone!

Joffan Feb22-12 09:27 PM

Re: Help Evaluating Logarithm
 
Also, if you didn't want to change base,
[tex]
\begin{align}
log_34 \times log_29 & = log_32^2 \times log_23^2\\
&= 2log_32 \times 2log_23\\
&= 4(log_32.log_23)\\
&= 4(log_33) \\
&=4
\end{align}
[/tex]


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