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I am trying to prove that: $$y^z.z^y=z^x.x^z=x^y.y^x$$

Given that: $${{x(y+z-x)}\over {log(x)}}={{y(z+x-y)} \over {log(y)}}={{z(x+y-z)} \over{log(z)}}$$

I started to solve it as follows: $${{x(y+z-x)}\over {log(x)}}={{y(z+x-y)} \over {log(y)}}={{z(x+y-z)} \over{log(z)}}={1\over k} \tag{1}$$ From here I did some minor manipulations and got $$ {y.log(x)}=(y+z-x)k$$ $$\implies log(x^y)=k(y+z-x)\tag{2} $$

Similarly, $$log(y^x)=k(z+x-y) \tag{3} $$ $$(2)+(3)$$

$$log(x^y.y^x)=k(2z) \tag{4}$$ Again Using (1) and manipulating like before we end up with

$$log(z^x.x^z)=k(2y) \tag{5}$$ $$log (y^z.z^y)=k(2x) \tag{6}$$ How do I proceed from here? If I put it such that equations (4),(5)and(6) have only$1 \over k$ on the RHS and then equate them I'd still have other terms in the denominator.If it were not for the remaining terms I could've taken their anti-logs. Is there a better way to solve this?

Infinity 207
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1 Answers1

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You can easily solve this very elementary:

Using $\log$ on the equation gives us $$z\log(y) + y\log(z) = x\log(z) + z\log(x) = x\log(y) + y\log(x)$$

Multiplying with $${{x(y+z-x)}\over {\log(x)}}={{y(z+x-y)} \over {\log(y)}}={{z(x+y-z)} \over{\log(z)}}$$ leads to \begin{align*}&\;zy(z+x-y) + yz(x+y-z) \\=&\;xz(x+y-z) + zx(y+z-x) \\=&\;xy(z+x-y) + yx(y+z-x)\end{align*}

which simplyfies to $$\iff 2xyz = 2xyz = 2xyz$$ and we are done.

And because all modifications were equivalent the initial equation holds.

Gono
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