I'm trying to bring this expression:

$$\frac{5}{6}\log\left(\frac{5}{4}\right) - \frac{1}{6}\log(2)$$

To this one:

$$\log\left(\frac{5}{4}\right) - \frac{1}{6}\log\left(\frac{5}{2}\right)$$

Where $\log$ is the natural algorithm.

I know the two expressions are equal (checked with wolfram) but I really can't find the correct passages... Could you please help me?

up vote 6 down vote accepted


$= \log\left(\frac54\right)-(\frac16\log\left(\frac54\right)+\frac16\log(2))$



I should add: Subtracting logs is equivalent to division. The reason for potential problems here is $\frac{\frac{a}{b}}c = \frac{a}{b\cdot c}$ and not $\frac{a}{(\frac{b}{c})}$

Hint: Begin with$$\frac56\log\left(\frac54\right)-\frac16\log(2)=\log\left(\frac54\right)-\frac16\log\left(\frac54\right)-\frac16\log(2).$$

  • Ty very much for the hint, I was on the right road, but missed the last passage that, imho, Phil stated very simply and clear – Alessio Martorana Sep 22 at 0:25

We have

$$\frac{5}{6}\log\left(\frac{5}{4}\right) - \frac{1}{6}\log 2=\overbrace{\frac{5}{6}\log\left(\frac{5}{4}\right)+\color{red}{\frac{1}{6}\log\left(\frac{5}{4}\right)}} -\overbrace{\color{red}{\frac{1}{6}\log\left(\frac{5}{4}\right)}- \frac{1}{6}\log 2}=$$

$$=\log\left(\frac{5}{4}\right)-\frac16 \log\left(\frac{5}{2}\right)$$

indeed recall that $\log A+ \log B= \log AB$ and therefore

$$-\frac{1}{6}\log\left(\frac{5}{4}\right)- \frac{1}{6}\log 2=-\frac16\left[\log\left(\frac{5}{4}\right)+\log 2\right]=-\frac16\log\left(\frac{5}{4}\cdot 2\right)=-\frac16\log\left(\frac{5}{2}\right)$$

Alt. hint:   by brute force, using only $\,\log \frac{a}{b} = \log a - \log b\,$ and $\,\log a^n = n \log a\,$:

$$\small \frac{5}{6}\log\frac{5}{4} - \frac{1}{6}\log 2 = \frac{1}{6}\left(5 \log 5 - 5 \log 4-\log 2\right) = \frac{1}{6}\left(5 \log 5 - 10 \log 2-\log 2\right) = \frac{1}{6}\left(5 \log 5 - 11 \log 2\right) $$

Now do the same for $\,\log \frac{5}{4} - \frac{1}{6}\log \frac{5}{2}\,$ and compare.

$$\frac{\color{blue}{5}}{6}\log \frac{5}{4} - \frac{1}{6}\log 2 = \frac{\color{blue}{6-1}}{6}\log\frac{5}{4} - \frac{1}{6}\log2 = \log \frac{5}{4} - \frac{1}{6}\left(\color{green}{\log\frac{5}{4} + \log 2 }\right)$$$$ = \log \frac{5}{4} - \frac{1}{6}\color{green}{ \log \left( \frac{5}{4} \cdot 2 \right)} = \log \frac{5}{4} - \frac{1}{6} \log \frac{5}{2} $$

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