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The question is to find out the solution of the following system of equations $$\log_{10}(x^3-x^2)=\log_5y^2$$ $$\log_{10}(y^3-y^2)=\log_5z^2$$ $$\log_{10}(z^3-z^2)=\log_5x^2$$

Letting $f(x)=5^{\log_{10}(x^3-x^2)}$.The equations can be expressed as $f(x)=y^2$,$f(y)=z^2$,$f(z)=x^2$.Now $(x^3-x^2)=x^2(x-1)$.So that it is a increasing in nature.Hence if $x>y$ then $f(x)>f(y)\implies y>z\implies z>x$.A contradiction.Hence the only solution is $x=y=z$.Is there another way to solve this problem without going in for contradiction?Thanks.

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    It seems there is not – 2017-02-22

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$x,y,z>1$

With the equation system we get

$\displaystyle g(x):=( (x^3-x^2)^{1.5\log_{10}5} - (x^3-x^2)^{\log_{10}5} )^ {1.5\log_{10}5} - $

$\hspace{1.2cm}\displaystyle -( (x^3-x^2)^{1.5\log_{10}5} - (x^3-x^2)^ {\log_{10}5})^{ \log_{10}5} - x^{2\log_5 10}=0$

which makes with real $\enspace x\enspace $ only sense for $\enspace x\geq x_0\enspace $ with

$\displaystyle x_0:=\frac{1}{3}\left(1+2^{-\frac{1}{3}}\left( \sqrt[3]{29-3\sqrt{93}}+\sqrt[3]{29+3\sqrt{93}} \right)\right)\approx 1.465571231876768...$ .

It follows $\enspace g(x)<0\enspace $ for $\enspace x<5$, $\enspace g(x)>0\enspace $ for $\enspace x>5$, $\enspace g(5)=0$

and therefore the only solution of the equation system is $\enspace x=y=z=5$ .