simplify this log expression

**differentiate** · March 28th 2010, 06:42 PM

This is what I had:

$\log _354 + log_35 - log_84$
$log_33 + log_33 + log_32 + log_35 - log_8(8^\frac{2}{3})$
$2\tfrac{1}{3} + log_310$

is this correct? can I simplify any further?

thanks

**bigwave** · March 28th 2010, 06:52 PM

$\log_a(MN) = \log_aM+\log_aN<br />$

**Prove It** · March 28th 2010, 06:52 PM

Originally Posted by differentiate

This is what I had:

$\log _354 + log_35 - log_84$
$log_33 + log_33 + log_32 + log_35 - log_8(8^\frac{2}{3})$
$2\tfrac{1}{3} + log_310$

is this correct? can I simplify any further?

thanks

$\log_3{54} + \log_3{5} - \log_8{4}$

$= \log_3{(27 \cdot 2)} + \log_3{5} - \log_8{\left(8^{\frac{2}{3}}\right)}$

$= \log_3{27} + \log_3{2} + \log_3{5} - \frac{2}{3}$

$= \log_3(3^3) + \log_3{(2\cdot 5)} - \frac{2}{5}$

$= 3 + \log_3{10} - \frac{2}{5}$

$= \frac{13}{5} + \log_3{10}$ .

So yes, you are correct. And no, it can't be simplified any further. The best you could do is to convert it to a base 10 or base $e$ .

**differentiate** · March 28th 2010, 07:06 PM

thanks a bunch Prove it!

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