How is this logarithm problem solved?

Question

If $\dfrac{xy\log xy}{x+y} = \dfrac{yz\log yz}{y+z} = \dfrac{zx\log zx}{z+x}$ show that $x^x = y^y = z^z$.

I tried equating to a constant $k$ and adding up

I also tried adding up the numerators and denominators to find each ratio

But still i am not able to figure out how to reach the final result

Does it should be $\cdots=\displaystyle\frac{zx\log zx}{z+\color{red}{x}}$? — Solumilkyu, Apr 22 '16 at 09:51

score 4 · Accepted Answer · 2016-04-22T10:49:25.577

4

Rewrite

$$\frac{\log(x)+\log(y)}{\frac1x+\frac1y}= \frac{\log(y)+\log(z)}{\frac1y+\frac1z}= \frac{\log(z)+\log(x)}{\frac1z+\frac1x}.$$

Then by the rules of proportions,

$$\frac{\log(x)}{\frac1x}=\frac{\log(y)}{\frac1y}=\frac{\log(z)}{\frac1z}.$$

edited Apr 22 '16 at 10:49

answered Apr 22 '16 at 09:56

Should your last term have been $\dfrac{\log z}{\frac 1 z}$? $\qquad$ – Michael Hardy Apr 22 '16 at 10:06
Can i do it by equating the first equation to k – sidt36 Apr 22 '16 at 10:07
Which proportion rule is used – sidt36 Apr 22 '16 at 10:27
@MichaelHardy: was called on the phone before finishing :) – Apr 22 '16 at 10:49

1 Answers1