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)) 
July 8th, 2017, 02:29 PM  #3  
Senior Member Joined: May 2016 From: USA Posts: 679 Thanks: 283  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 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! 

Tags 
logs 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Having some problems understanding logs (natural logs)  leo255  Algebra  4  January 26th, 2014 09:49 AM 
Logs Help  Lewan  Algebra  3  May 7th, 2013 12:13 PM 
logs  milanstar  Algebra  3  May 12th, 2012 12:13 PM 
Help on solving problems with logs and natural logs!  rokr32  Algebra  1  February 1st, 2012 04:13 PM 
logs  milanstar  Calculus  0  January 1st, 1970 12:00 AM 