Mastering Logarithms: Simplifying Logarithmic Expressions

[theintarnets]

Homework Statement

Evaluate without a calculator:
(log₃4 + log₂9)² - (log₃4 - log₂9)²

Homework Equations

The Attempt at a Solution

(log₃4 + log₂9)² - (log₃4 - log₂9)²

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

And now I'm stuck...

 

[rock.freak667]

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]

$$a^2-b^2=(a-b)(a+b)$$EDIT: how to delete message

 

[eumyang]

(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]

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

 

[Joffan]

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 log_nx^k = k.log_nx
and that log_ab $\times$ log_bc = log_ac

 

[theintarnets]

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]

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]

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]

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]

Okie. Trying again now.

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

 

[eumyang]

Show us what you have so far.

 

[theintarnets]

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]

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]

Yay! Thanks so much everyone!

 

[Joffan]

Also, if you didn't want to change base,
$$\begin{align} log_34 \times log_29 & = log_32^2 \times log_23^2\\ &= 2log_32 \times 2log_23\\ &= 4(log_32.log_23)\\ &= 4(log_33) \\ &=4 \end{align}$$