4
$\begingroup$

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?

$\endgroup$
8
$\begingroup$

$\frac56\log\left(\frac54\right)-\frac16\log(2)=\log\left(\frac54\right)-\frac16\log\left(\frac54\right)-\frac16\log(2)$

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

$=\log\left(\frac54\right)-\frac16\log\left(\frac{5\cdot2}{4}\right)$

$=\log\left(\frac54\right)-\frac16\log\left(\frac52\right)$

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 $\frac{a}{(\frac{b}{c})}$

$\endgroup$
6
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Alessio Martorana Sep 22 '18 at 0:25
4
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.