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

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- Nov 23rd 2012, 04:18 AMTutuplwase 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 AMchiroRe: 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 PMTutuRe: plwase help me check my answer!
Thanks! I found.the answer key, it says its.lg2..

- Nov 23rd 2012, 06:04 PMSorobanRe: 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 PMtopsquarkRe: plwase help me check my answer!
- Nov 24th 2012, 06:44 AMtopsquarkRe: plwase help me check my answer!