One logarithm question

**BG5965** · November 15th 2008, 06:00 AM

Hi, I am having problem with one difficult looking logarithm. Please help me!
$ 3log_2 \frac{9}{8} + 4*log_2 \frac{16}{15} - 2 log_2 \frac{24}{25} $

Thanks! Please show your working -

*EDIT: This is supposed to be solved without an calculator, so can you please simplify it down to something which you can easily work with, with logarithm laws.

Thanks

**Rapha** · November 15th 2008, 06:12 AM

Hey

Originally Posted by BG5965

Hi, I am having problem with one difficult looking logarithm. Please help me!

Thanks! Please show your working.

this is really hard to read, man

$3log_2 \frac{9}{8} + 4*log_2 \frac{16}{15} - 2 log_2 \frac{24}{25}$

it is $log_a(b) = ln(b) /ln (a)$

therefor

$1/ln(2) [ 3 ln(\frac{9}{8}) + 4*ln(\frac{16}{15}) - 2 ln(\frac{24}{25})] = 0.562$

**BG5965** · November 15th 2008, 03:08 PM

Thanks, but I wouldn't be able to do that without an calculator. Is there another way to simplify it down to basic logarithms that we can easily use?
Thanks

**mr fantastic** · November 15th 2008, 03:55 PM

Originally Posted by BG5965

Hi, I am having problem with one difficult looking logarithm. Please help me!
$ 3log_2 \frac{9}{8} + 4*log_2 \frac{16}{15} - 2 log_2 \frac{24}{25} $

Thanks! Please show your working -

*EDIT: This is supposed to be solved without an calculator, so can you please simplify it down to something which you can easily work with, with logarithm laws.

Thanks

From the usual log rules:

$= 3 (\log_2 9 - \log_2 8) + 4 (\log_2 16 - \log_2 15) - 2 (\log_2 24 - \log_2 25)$

$= 3 (\log_2 3^2 - \log_2 2^3) + 4 (\log_2 2^4 - \log_2 (3 \cdot 5)) - 2 (\log_2 (3 \cdot 2^3) - \log_2 5^2)$

From the usual log rules:

$= 3 (2 \log_2 3 - 3) + 4 (4 - (\log_2 3 + \log_2 5) ) - 2 (\log_2 3 + \log_2 2^3 - 2 \log_2 5)$

$= 3 (2 \log_2 3 - 3) + 4 (4 - \log_2 3 - \log_2 5 ) - 2 (\log_2 3 + 3 - 2 \log_2 5)$

$= 6 \log_2 3 - 9 + 16 - 4 \log_2 3 - 4 \log_2 5 - 2 \log_2 3 -6 + 4 \log_2 5$

and the mopping up is left for you to do.

I get 1 as the answer.

But you should check all my work carefully because I am well known for making careless mistakes.

**BG5965** · November 15th 2008, 04:10 PM

Thanks - I just realized logarithms can sometime be just 'complex' algebra.
Anyway, I canceled out all the like terms (the logarithms with the common base) and was left with -9 + 10 = 1

I think you were right.

Thanks for your help.

**Soroban** · November 15th 2008, 06:28 PM

Hello, BG5965!

No wonder you're having problems . . . this is a messy one!

And I agree with Mr. F . . .

$3\log_2\left(\tfrac{9}{8}\right) + 4\log_2\left(\tfrac{16}{15}\right) - 2\log_2\left(\tfrac{24}{25}\right)$

I'll work on the three terms separately, and combine them at the end.

$3\log_2\left(\frac{9}{8}\right) \;=\;3\bigg[\log_2(9) - \log_2(8)\bigg] \;=\;3\bigg[\log_2(3^2) - \log_2(2^3)\bigg]$

. . $=\;3\bigg[2\log_2(3) - 3\bigg] \;=\;\boxed{6\log_2(3) - 9}$

$4\log_2\left(\frac{16}{15}\right) \;=\;4\bigg[\log_2(16) - \log_2(15)\bigg] \;=\;4\bigg[\log_2(2^4) - \log_2(3\cdot5)\bigg]$

. . $= \;4\bigg[4 - \bigg(\log_2(3) + \log_2(5)\bigg)\bigg] \;=\;\boxed{16 - 4\log_2(3) - 4\log_2(5)}$

$-2\log_2\left(\frac{24}{25}\right) \;=\;-2\bigg[\log_2(24) - \log_2(25)\bigg] \;=\;-2\bigg[\log_2(2^3\cdot3) - \log_2(5^2)\bigg]$

. . $= \;-2\bigg[\log_2(2^3) + \log_2(3) - 2\log_2(5)\bigg] \;=\;-2\bigg[3 + \log_2(3) - 2\log_2(5)\bigg]$

. . $= \;\boxed{-6 - 2\log_2(3) + 4\log_2(5)}$

Combine the three expressions:

. . $\begin{array}{ccccccc}6\log_2(3) & & & - & 9 \\ \text{-}4\log_2(3) & - & 4\log_2(5) & + & 16 \\ \text{-}2\log_2(3) &+& 4\log_2(5) &-& 6 \\ \hline \\[-4mm] & & & &\boxed{{\color{red}1}} \end{array}$

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