Alg9.4.25

If $a=1+\log_x(yz),b=1+\log_y(zx),c=1+\log_z(xy)\,$ then show that ab+bc+ca=abc.

$1+\log_x(yz)=a\,$

$\log_x(yz)=a-1\,$

$\log yz=x^{a-1}\,$

$yz=x^{a-1}\,$

$xyz=x^a\,$

Hence

$x=(xyz)^{\frac{1}{a}}\,$

$y=(xyz)^{\frac{1}{b}}\,$

$z=(xyz)^{\frac{1}{c}}\,$

Multiplying the above three, we get

$xyz=(xyz)^{\frac{1}{a}}\cdot(xyz)^{\frac{1}{b}}\cdot(xyz)^{\frac{1}{c}}\,$

Simplifying

$xyz=(xyz)^{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\,$

Comparing the indices on both sides,we get

$1=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\,$

Simplifying further

$abc=ab+bc+ca\,$

Hence proved.