# plwase help me check my answer!

• Nov 23rd 2012, 04:18 AM
Tutu
plwase help me check my answer!
I do.not have.the.answer.to this question, could.I.check.mine with yours?lg(75/16)-2lg(5/9)+lg(32/343), I got lg(486/343)...if I had gotten it wrong(most likely) could you please show me the working? Thankvyou so much
• Nov 23rd 2012, 08:05 AM
chiro
Re: plwase help me check my answer!
Hey Tut.

Doing the working out I get log(75/16) - 2log(5/9) + log(32/343) = log([75/16]/(25/81)*(32/343)) = log(194400/137200)

Here is the working out from R.
> 75*81*32
[1] 194400
> 16*25*343
[1] 137200
• Nov 23rd 2012, 04:24 PM
Tutu
Re: plwase help me check my answer!
Thanks! I found.the answer key, it says its.lg2..
• Nov 23rd 2012, 06:04 PM
Soroban
Re: plwase help me check my answer!
Hello, Tutu!

You gave us a problem with a typo . . .

Quote:

Simplify: .log(75/16) - 2log(5/9) + log(32/243)

We have:
. . $\displaystyle \log\left(\frac{75}{16}\right) - \log\left(\frac{5}{9}\right)^2 + \log\left(\frac{32}{243}\right)$

. . . . $\displaystyle =\;\log\left(\frac{75}{16}\right) - \log\left(\frac{25}{81}\right) + \log\left(\frac{32}{243}\right)$

. . . . $\displaystyle =\;\log\left(\frac{75}{16} \div \frac{25}{81} \times \frac{32}{243}\right)$

. . . . $\displaystyle =\;\log\left(\frac{{\color{blue}\rlap{//}}75^3}{{\color{green}\rlap{//}}16}\cdot\frac{{\color{red}\rlap{//}}81}{{\color{blue}\rlap{//}}25}\cdot \frac{{\color{green}\rlap{//}}32^2}{{\color{red}\rlap{///}}243_3}\right)$

. . . . $\displaystyle =\;\log(2)$
• Nov 23rd 2012, 06:12 PM
topsquark
Re: plwase help me check my answer!
Quote:

Originally Posted by Tutu
Thanks! I found.the answer key, it says its.lg2..

The answer key is wrong. Both of you have the correct solution, just written in a different form. (My answer is in a third form.) lg(2) can't be right because log(2) + 5 lg(3) - 3 log(7) does the job.

-Dan
• Nov 24th 2012, 06:44 AM
topsquark
Re: plwase help me check my answer!
Quote:

Originally Posted by topsquark
The answer key is wrong. Both of you have the correct solution, just written in a different form. (My answer is in a third form.) lg(2) can't be right because log(2) + 5 lg(3) - 3 log(7) does the job.

-Dan

Hmm....Seems I can't edit my post for some reason. Anyway, thanks for the catch Soroban. As always, nice job! (And I didn't see your solution before today.)

-Dan