Given: $\frac{\log_{10}z}{y-x} = \frac{\log_{10}y}{x-z} = \frac{\log_{10}x}{z-y},$ prove: a) $x^x\times y^y\times z^z =1,$ b) $x\times y \times z =1.$

Question

Given: $ \frac{\log_{10}z}{y-x} = \frac{\log_{10}y}{x-z} = \frac{\log_{10}x}{z-y} $

Prove: $ a)\ x^x\times y^y\times z^z =1 \\ b) \ x\times y \times z =1 $

Here is a few step I tried to solve: $$ \log_{10}z^zy^yx^x => z\log_{10}z + y\log_{10}y + x\log_{10}x $$ and find each $\log_{10}z, \log_{10}y \ and \log_{10}x$ in the form of one given the equation by equalizing the denominator:

$(x-z)(y-x)\log_{10}z = (y-x)(z-y)\log_{10}y = (y-x)(x-z)log_{10}x $ and calculating $ \frac{\log_{10}z }{\log_{10}y}, \frac{\log_{10}z }{\log_{10}x} $

Somehow I couldn't figure out the how to get $ x^x\times y^y\times z^z =1... $

Can you help me to prove the equation? Thank you.

Answer 1

Let $k$ be equal to our fractions: $$ \frac{\log_{10}z}{y-x} = \frac{\log_{10}y}{x-z} = \frac{\log_{10}x}{z-y}=k.$$

Thus, $$\prod_{cyc}z=\prod_{cyc}10^{k(y-x)}=10^0=1.$$

Also, $$\prod_{cyc}z^z=\prod_{cyc}10^{kz(y-x)}=10^0=1.$$ I used $$\sum_{cyc}k(y-x)=k(y-x+z-y+x-z)=0$$ and $$\sum_{cyc}kz(y-x)=k(zy-zx+xz-xy+yx-yz)=0.$$

Answer 2

Here the common base is 10, then let all tharatios b equal to $l$, then $$\frac{\log x}{y-z}=\frac{\log y}{z-x}=\frac{\log z}{x-y}=k \implies x=10^{k(y-z)}, y=10^{k(z-x)}, z=10^{k(x-y)}.$$ $$F=x^x. y^y. z^z=10^{k(xy-xz)}.10^{k(yz-yx)}. 10^{(zx-zy)}=10^0=1$$ $$G=x.y.z=10^{k(y-z)}.10^{k(z-x)}.10^{k(x-y)}=10^0=1.$$