# Combining log terms

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

• Not quite. The $c^5$ belongs in the denominator. – Mark Viola Apr 17 '15 at 4:44
• $c$ ought to be in the denominator. – zahbaz Apr 17 '15 at 4:44
• Thank you but why? isnt it the form of $a/b/c$ which equals $ac/b$ ? – Max Payne Apr 17 '15 at 4:47
• 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. – zahbaz Apr 17 '15 at 4:50
• Ah! This might be your error... $\ln a - \ln b -\ln c \ne \ln a - \ln\frac{b}{c}$. Why? Consider: $\ln a - (\ln b + \ln c) = \ln a - \ln bc = \ln\frac{a}{bc}$ – zahbaz Apr 17 '15 at 4:55

$$\dfrac{(2\ln a- \ln b - 5\ln c)}{2}$$ $$=\dfrac{(\ln a^2- \ln b - \ln c^5)}{2}$$ $$= \dfrac{\left(\ln \dfrac{a^2}{b} - \ln c^5\right)}{2}$$ $$= \dfrac{1}{2}\ln \dfrac{a^2}{bc^5}$$ $$= \ln \dfrac{a}{\sqrt{bc^5}}$$