# If (xy log xy)/(x+y) = (yz log yz)/(y +z)= (zx log zx)/(z + x) Show that x^x = y^y = z^z?

Jul 15, 2017

#### Explanation:

$\frac{x y \log x y}{x + y} = \frac{y z \log y z}{y + z} = \frac{z x \log z x}{z + x}$

$\implies \frac{\log x y}{\frac{x + y}{x y}} = \frac{\log y z}{\frac{y + z}{y z}} = \frac{\log z x}{\frac{z + x}{z x}}$

$\implies \frac{\log x + \log y}{\frac{1}{x} + \frac{1}{y}} = \frac{\log y + \log z}{\frac{1}{y} + \frac{1}{z}} = \frac{\log z + \log x}{\frac{1}{z} + \frac{1}{x}} = k$ say

then $k \left(\frac{1}{x} + \frac{1}{y}\right) = \log x + \log y$ .........................(1)

$k \left(\frac{1}{y} + \frac{1}{z}\right) = \log y + \log z$ .........................(2) and

$k \left(\frac{1}{z} + \frac{1}{x}\right) = \log z + \log x$ .........................(3)

Adding the three, we get $2 k \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) = 2 \left(\log x + \log y + \log z\right)$ or

$k \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) = \log x + \log y + \log z$ .........................(4)

Now subtracting (1), (2) and (3) from (4), we get

$\frac{k}{z} = \log z$ i.e. $k = z \log z = \log {z}^{z}$ .........................(5)

$\frac{k}{x} = \log x$ i.e. $k = x \log x = \log {x}^{x}$ .........................(6)

$\frac{k}{y} = \log y$ i.e. $k = y \log y = \log {y}^{y}$ .........................(5)

(5), (6) and (7) give us

$\log {x}^{x} = \log {y}^{y} = \log {z}^{z}$

i.e. ${x}^{x} = {y}^{y} = {z}^{z}$