# Help Evaluating Logarithm

1. ### theintarnets

64
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....

Last edited: Feb 22, 2012
2. ### rock.freak667

6,221
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).

3. ### Karamata

60
$$a^2-b^2=(a-b)(a+b)$$

EDIT: how to delete message

4. ### eumyang

1,347
It also looks like you went from
(log34 + log29)2
to
(log342 + log292),
which is not true.

5. ### theintarnets

64
Ohhhhh, I see. I shall re-attempt now. Thanks.

6. ### Joffan

472
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 $\times$ logbc = logac

7. ### theintarnets

64
Erm...Nevermind. I'm still stuck. Can someone walk me through it please?

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

### Staff: Mentor

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

9. ### theintarnets

64
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...

10. ### eumyang

1,347
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.

11. ### theintarnets

64
Okie. Trying again now.

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

Last edited: Feb 22, 2012
12. ### eumyang

1,347
Show us what you have so far.

13. ### theintarnets

64
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.

14. ### eumyang

1,347
That's correct.

15. ### theintarnets

64
Yay!!! Thanks so much everyone!

16. ### Joffan

472
Also, if you didn't want to change base,
\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}