# logs again

• Jan 5th 2010, 02:48 AM
logs again
simplify log(base4/9)(27/16) without a calculator

i've tried everything i could think of without a calculator, so here i am.
• Jan 5th 2010, 03:00 AM
Prove It
Quote:

simplify log(base4/9)(27/16) without a calculator

i've tried everything i could think of without a calculator, so here i am.

$\displaystyle \frac{27}{16} = \frac{3}{4}\cdot\frac{9}{4}$

$\displaystyle = \frac{3}{4}\left(\frac{4}{9}\right)^{-1}$.

So $\displaystyle \log_{\frac{4}{9}}\frac{27}{16} = \log_{\frac{4}{9}}\left[\frac{3}{4}\left(\frac{4}{9}\right)^{-1}\right]$

$\displaystyle = \log_{\frac{4}{9}}\frac{3}{4} + \log_{\frac{4}{9}}\left(\frac{4}{9}\right)^{-1}$

$\displaystyle = \log_{\frac{4}{9}}\frac{3}{4} - 1$

$\displaystyle = \log_{\frac{4}{9}}{3} - \log_{\frac{4}{9}}{4} - 1$.
• Jan 5th 2010, 03:06 AM
Quote:

Originally Posted by Prove It
$\displaystyle \frac{27}{16} = \frac{3}{4}\cdot\frac{9}{4}$

$\displaystyle = \frac{3}{4}\left(\frac{4}{9}\right)^{-1}$.

So $\displaystyle \log_{\frac{4}{9}}\frac{27}{16} = \log_{\frac{4}{9}}\left[\frac{3}{4}\left(\frac{4}{9}\right)^{-1}\right]$

$\displaystyle = \log_{\frac{4}{9}}\frac{3}{4} + \log_{\frac{4}{9}}\left(\frac{4}{9}\right)^{-1}$

$\displaystyle = \log_{\frac{4}{9}}\frac{3}{4} - 1$

$\displaystyle = \log_{\frac{4}{9}}{3} - \log_{\frac{4}{9}}{4} - 1$.

silly me, i was trapped in the thought of solving it! thanks for the enlightenment on my silly misunderstanding
• Jan 5th 2010, 06:07 AM
Quote:

simplify log(base4/9)(27/16) without a calculator

i've tried everything i could think of without a calculator, so here i am.

Here's an alternative approach.

Let $\displaystyle x = \log_{\frac49}\left(\frac{27}{16}\right)$

Then $\displaystyle \left(\frac49\right)^x = \frac{27}{16}$

$\displaystyle \Rightarrow \left(\left(\frac23\right)^2\right)^x = \frac{3^3}{2^4}$

$\displaystyle \Rightarrow \left(\frac23\right)^{2x}=\frac12\left(\frac23\rig ht)^{-3}$

$\displaystyle \Rightarrow 2x = \log_{\frac23}\left(\frac12\right)-3\log_{\frac23}\left(\frac23\right)$
$\displaystyle =-\log_{\frac23}(2)-3$
$\displaystyle \Rightarrow x = -\tfrac12(\log_{\frac23}(2)+3)$

...but whether you think this answer is any simpler is a matter for debate!

If you want to get rid of fractional bases, you can express this in terms of $\displaystyle \log_2$ using the formula
$\displaystyle \log_{\frac23}(2) = \frac{\log_22}{\log_2(\tfrac23)}$
$\displaystyle =\frac{1}{1-\log_23}$
which then gives:
$\displaystyle x = \frac{4-3\log_23}{2(\log_23-1)}$
but I don't think you can rid of logs altogether!