My Math Forum Simplifying Logarithmic Functions

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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 $$\ln(8x)-\ln(4x^{2})+\ln(8x^{6}) = \ln(8x)+\ln(8x^{6}) -\ln(4x^{2}) = \ln(8x)+\ln(2x^{4})$$ Thanks from secsci 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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