Hello everyone, I had a question about a problem involving some logs:

$\displaystyle 5log_32+2log_910=?$

I'm not quite sure where to start with this problem, any suggestions would be appreciated.

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- Oct 12th 2010, 07:29 AMrtblueLogarithms
Hello everyone, I had a question about a problem involving some logs:

$\displaystyle 5log_32+2log_910=?$

I'm not quite sure where to start with this problem, any suggestions would be appreciated. - Oct 12th 2010, 07:36 AMTheEmptySet
A good place to start might be to use the change of base formula

Logarithm - Wikipedia, the free encyclopedia

$\displaystyle \log_9(a)=\frac{\log_{3}(a)}{\log_3(9)}=...$

You will need to use other log properties as well. - Oct 12th 2010, 07:38 AMPlato
Hint: $\displaystyle 2\log_9(10)=\log_3(10)$.

- Oct 12th 2010, 11:34 AMHallsofIvy
That's because $\displaystyle y= log_9(x)$ is the same as $\displaystyle x= 9^y= (3^2)^y= 3^{2y}$ so that $\displaystyle log_3(x)= 2y= 2log_9(x)$.

- Oct 12th 2010, 12:21 PMbigwaveuse ln
using as has been mentioned you could too rewrite to:

$\displaystyle \log_3{32} + \log_9{100}$

then

$\displaystyle \frac{ln32}{ln3} + \frac{ln100}{ln9}

\Rightarrow

3.15 + 4.19 = 5.25$(approx)