# One logarithm question

• Nov 15th 2008, 06:00 AM
BG5965
One logarithm question*EDITED
$\displaystyle 3log_2 \frac{9}{8} + 4*log_2 \frac{16}{15} - 2 log_2 \frac{24}{25}$

*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
• Nov 15th 2008, 06:12 AM
Rapha
Hey

Quote:

Originally Posted by BG5965
http://i301.photobucket.com/albums/nn74/BL5965/5d.jpg

this is really hard to read, man

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

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

therefor

$\displaystyle 1/ln(2) [ 3 ln(\frac{9}{8}) + 4*ln(\frac{16}{15}) - 2 ln(\frac{24}{25})] = 0.562$
• Nov 15th 2008, 03:08 PM
BG5965
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
• Nov 15th 2008, 03:55 PM
mr fantastic
Quote:

Originally Posted by BG5965
$\displaystyle 3log_2 \frac{9}{8} + 4*log_2 \frac{16}{15} - 2 log_2 \frac{24}{25}$

*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:

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

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

From the usual log rules:

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

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

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

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.
• Nov 15th 2008, 04:10 PM
BG5965
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.

• Nov 15th 2008, 06:28 PM
Soroban
Hello, BG5965!

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

And I agree with Mr. F . . .

Quote:

$\displaystyle 3\log_2\left(\tfrac{9}{8}\right) + 4\log_2\left(\tfrac{16}{15}\right) - 2\log_2\left(\tfrac{24}{25}\right)$
I'll work on the three terms separately, and combine them at the end.

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

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

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

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

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

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

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

Combine the three expressions:

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