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 July 8th, 2017, 01:24 PM #1 Newbie   Joined: Jul 2017 From: United Kingdom Posts: 4 Thanks: 0 Math Focus: General Logs I have a little problem with Logs. (log(3x)+log(9x^2))/3 I worked it out to be Log(27x) but it can be simplified further down to Log(3x) Can anybody help me figure out how? Thanks!
 July 8th, 2017, 01:53 PM #2 Newbie   Joined: May 2017 From: California Posts: 11 Thanks: 1 Try to use log(3x) + log(9x^2) = log(3x*(9x^2)) then use log(x) * a = log(x^a).. so log(x) / a = log(x^(1/a)) Thanks from JayWalk
July 8th, 2017, 02:29 PM   #3
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Quote:
 Originally Posted by JayWalk I have a little problem with Logs. (log(3x)+log(9x^2))/3 I worked it out to be Log(27x) but it can be simplified further down to Log(3x) Can anybody help me figure out how? Thanks!
$\dfrac{\log(3x) + \log(9x^2)}{3} = \dfrac{\log(3x) + \log(\{3x\}^2)}{3} =$

$\dfrac{\log(3x) + 2\log(3x)}{3} = \dfrac{3\log(3x)}{3} = \log(3x).$

EDIT: This is just a different way from in the previous post.

$\dfrac{\log(3x) + \log(9x^2)}{3} = \dfrac{\log(3x * 9x^2)}{3} = \dfrac{\log(27x^3)}{3}.$

Now you can go

$\dfrac{\log(27x^3)}{3} = \dfrac{\log(\{3x\}^3)}{3} =$

$\dfrac{3\log(3x)}{3} = \log(3x).$

Or you can go:

$\dfrac{\log(27x^3)}{3} = \dfrac{1}{3} * \log(27x^3) =$

$\log(\{27x^3\}^{(1/3)}) = \log \left ( \sqrt[3]{27x^3} \right ) = \log(3x).$

Lots of ways to proceed.

Last edited by skipjack; July 8th, 2017 at 09:14 PM.

 July 8th, 2017, 05:23 PM #4 Newbie   Joined: Jul 2017 From: United Kingdom Posts: 4 Thanks: 0 Math Focus: General I can't believe I didn't see it at first. Thanks!

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