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April 28th, 2014, 07:31 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, 08:17 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 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 08:23 PM. 
April 28th, 2014, 08:39 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, 08:44 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 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, 08:46 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. 

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functions, logarithmic, simplifying 
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