If $\frac{x y log x y}{x + y} = \frac{y z log y z}{y + z} = \frac{z x log z x}{z + x}$ $(xy log xy)/(x+y) = (yz log yz)/(y +z)= (zx log zx)/(z + x)$ Show that $x^{x} = y^{y} = z^{z}$ $x^x = y^y = z^z$ ?

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If $\frac{x y log x y}{x + y} = \frac{y z log y z}{y + z} = \frac{z x log z x}{z + x}$ $(xy log xy)/(x+y) = (yz log yz)/(y +z)= (zx log zx)/(z + x)$ Show that $x^{x} = y^{y} = z^{z}$ $x^x = y^y = z^z$ ?

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Answer 1 · 2017-07-15T16:32:07

Please see below.

Explanation:

$\frac{x y log x y}{x + y} = \frac{y z log y z}{y + z} = \frac{z x log z x}{z + x}$ $(xylogxy)/(x+y)=(yzlogyz)/(y+z)=(zxlogzx)/(z+x)$

$\Rightarrow \frac{log x y}{\frac{x + y}{x y}} = \frac{log y z}{\frac{y + z}{y z}} = \frac{log z x}{\frac{z + x}{z x}}$ $=>(logxy)/((x+y)/(xy))=(logyz)/((y+z)/(yz))=(logzx)/((z+x)/(zx))$

$\Rightarrow \frac{log x + log y}{\frac{1}{x} + \frac{1}{y}} = \frac{log y + log z}{\frac{1}{y} + \frac{1}{z}} = \frac{log z + log x}{\frac{1}{z} + \frac{1}{x}} = k$ $=>(logx+logy)/(1/x+1/y)=(logy+logz)/(1/y+1/z)=(logz+logx)/(1/z+1/x)=k$ say

then $k (\frac{1}{x} + \frac{1}{y}) = log x + log y$ $k(1/x+1/y)=logx+logy$ .........................(1)

$k (\frac{1}{y} + \frac{1}{z}) = log y + log z$ $k(1/y+1/z)=logy+logz$ .........................(2) and

$k (\frac{1}{z} + \frac{1}{x}) = log z + log x$ $k(1/z+1/x)=logz+logx$ .........................(3)

Adding the three, we get $2 k (\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = 2 (log x + log y + log z)$ $2k(1/x+1/y+1/z)=2(logx+logy+logz)$ or

$k (\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = log x + log y + log z$ $k(1/x+1/y+1/z)=logx+logy+logz$ .........................(4)

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

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

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

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

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

${log x}^{x} = {log y}^{y} = {log z}^{z}$ $logx^x=logy^y=logz^z$

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

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If $\frac{x y log x y}{x + y} = \frac{y z log y z}{y + z} = \frac{z x log z x}{z + x}$ $(xy log xy)/(x+y) = (yz log yz)/(y +z)= (zx log zx)/(z + x)$ Show that $x^{x} = y^{y} = z^{z}$ $x^x = y^y = z^z$ ?

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If xylogxyx+y=yzlogyzy+z=zxlogzxz+x(xy log xy)/(x+y) = (yz log yz)/(y +z)= (zx log zx)/(z + x) Show that xx=yy=zzx^x = y^y = z^z?

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If $\frac{x y log x y}{x + y} = \frac{y z log y z}{y + z} = \frac{z x log z x}{z + x}$ $(xy log xy)/(x+y) = (yz log yz)/(y +z)= (zx log zx)/(z + x)$ Show that $x^{x} = y^{y} = z^{z}$ $x^x = y^y = z^z$ ?