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I have this particular problem. We have to combine the log terms into a single log term:

$$\dfrac{(2\ln a- \ln b - 5\ln c)}{2}$$

I did it in the following way :

$$''~= \ln a -\frac{1}{2}\ln b - \frac{5}{2} \ln c$$ $$= \ln\left(\left(\frac{a^2c^5}{b}\right)^{\frac{1}{2}}\right)$$

Is this correct approach?

I used the formula : $\log_ba-\log_bc=\log_b\left(\dfrac{a}{c}\right)$

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    $\begingroup$ Not quite. The $c^5$ belongs in the denominator. $\endgroup$ – Mark Viola Apr 17 '15 at 4:44
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    $\begingroup$ $c$ ought to be in the denominator. $\endgroup$ – zahbaz Apr 17 '15 at 4:44
  • $\begingroup$ Thank you but why? isnt it the form of $a/b/c$ which equals $ac/b$ ? $\endgroup$ – Max Payne Apr 17 '15 at 4:47
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    $\begingroup$ Try it with $-ln(x) = + ln(\frac{1}{x})$. And in general, as a visual aid, all the positives will go in the numerator, and the negatives in the denominator. $\endgroup$ – zahbaz Apr 17 '15 at 4:50
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    $\begingroup$ Ah! This might be your error... $\ln a - \ln b -\ln c \ne \ln a - \ln\frac{b}{c}$. Why? Consider: $\ln a - (\ln b + \ln c) = \ln a - \ln bc = \ln\frac{a}{bc}$ $\endgroup$ – zahbaz Apr 17 '15 at 4:55
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$$\dfrac{(2\ln a- \ln b - 5\ln c)}{2}$$ $$ =\dfrac{(\ln a^2- \ln b - \ln c^5)}{2} $$ $$= \dfrac{\left(\ln \dfrac{a^2}{b} - \ln c^5\right)}{2} $$ $$= \dfrac{1}{2}\ln \dfrac{a^2}{bc^5} $$ $$= \ln \dfrac{a}{\sqrt{bc^5}}$$

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  • $\begingroup$ Thank you very much for your comments and answer! This was really confusing! :) $\endgroup$ – Max Payne Apr 17 '15 at 4:57
  • $\begingroup$ is there a good resource (book etc) for learning logarithms? $\endgroup$ – Max Payne Apr 17 '15 at 5:08
  • $\begingroup$ You may want to dig into Khan Academy - lots of video tutorials. As for a book, try Precalculus by Young (Ch3 for exponential and logs). [1]: khanacademy.org/math/algebra2/logarithms-tutorial $\endgroup$ – zahbaz Apr 17 '15 at 5:21

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