
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
April 28th, 2014, 08:31 PM  #1 
Newbie Joined: Feb 2014 From: Australia Posts: 7 Thanks: 0  Simplifying Logarithmic Functions
Hi all, I am having issues with the simplification. It needs to be expressed as a single natural logarithm. $\displaystyle \ln(8x)\ln(2x)^{2}+\ln(2x^{2})^{3}$ The answer is $\displaystyle \ln(16x^{5})$ My Working: $\displaystyle \ln(8x)\ln(2x)^{2}+\ln(2x^{2})^{3}$ $\displaystyle \ln(8x)\ln(4x^{2})+\ln(8x^{6})$ $\displaystyle \ln(8x)\ln(4x^{2} \times 8x^{6})$ As per law and organisation I get to: $\displaystyle \ln\left [\frac{8x}{32x^{8}} \right ] $ I'm not sure if I'm going about it the right way as I have no idea to get to the answer. I've even considered cancelling out the 8$\displaystyle x$ but then theres still an exponent of 6 on the second 8$\displaystyle x$ Can someone shed the light on this for me so I can understand where I'm going wrong. Thanks in advance. 
April 28th, 2014, 09:17 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 Math Focus: Mainly analysis and algebra 
The problem you have is that $$\ln(8x)\ln(4x^{2})+\ln(8x^{6}) \ne \ln(8x)\ln(4x^{2} \times 8x^{6})$$ Instead $$\ln(8x)\ln(4x^{2})+\ln(8x^{6}) = \ln(8x)+\ln(8x^{6}) \ln(4x^{2}) = \ln(8x)+\ln(2x^{4})$$ Last edited by v8archie; April 28th, 2014 at 09:23 PM. 
April 28th, 2014, 09:39 PM  #3 
Newbie Joined: Feb 2014 From: Australia Posts: 7 Thanks: 0 
Hey Archie, I see exactly what you're saying, I didn't realise I was able to move the ln(4x^2) to the end of the expression. Thank you very much for your reply. 
April 28th, 2014, 09:44 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 Math Focus: Mainly analysis and algebra 
It's not so much about moving the terms around. The point is that you wrote $$\ln A + \ln B = \ln AB$$ But the minus sign doesn't operate on the $\ln B$. $$\ln A + \ln B = (\ln A  \ln B) = \ln \frac{A}{B} = \ln \frac{B}{A}$$ Or $$\ln A + \ln B = \ln \frac{1}{A} + \ln B = \ln \frac{B}{A}$$ 
April 28th, 2014, 09:46 PM  #5 
Newbie Joined: Feb 2014 From: Australia Posts: 7 Thanks: 0 
I see what you're saying, I made it seem like both terms operating off the minus when only 1 term was. I should have paid more attention to that. Appreciate you explaining this to me. 

Tags 
functions, logarithmic, simplifying 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Logarithmic Functions  strangepath  Algebra  2  April 13th, 2013 02:50 PM 
Exponential and Logarithmic Functions  Francois  Algebra  1  January 18th, 2013 09:25 AM 
differentiating logarithmic functions  mathkid  Calculus  2  December 7th, 2012 09:59 PM 
logarithmic functions  robasc  Algebra  1  July 21st, 2008 08:23 AM 
Logarithmic functions help.  RedBarron  Calculus  2  March 17th, 2008 05:53 PM 