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April 28th, 2014, 08:31 PM   #1
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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.
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April 28th, 2014, 09:17 PM   #2
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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 09:23 PM.
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April 28th, 2014, 09:39 PM   #3
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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.
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April 28th, 2014, 09:44 PM   #4
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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
$$-\ln A + \ln B = \ln \frac{1}{A} + \ln B = \ln \frac{B}{A}$$
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April 28th, 2014, 09:46 PM   #5
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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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