July 8th, 2017, 02:24 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, 02:53 PM  #2 
Newbie Joined: May 2017 From: California Posts: 12 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)) 
July 8th, 2017, 03:29 PM  #3  
Senior Member Joined: May 2016 From: USA Posts: 823 Thanks: 335  Quote:
$\dfrac{\log(3x) + 2\log(3x)}{3} = \dfrac{3\log(3x)}{3} = \log(3x).$ 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 10:14 PM.  
July 8th, 2017, 06:23 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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