Answer to Question #93602 in Classical Mechanics for Rinku Rai

Question

x=log bc base a, y=log ca base b,z=log ab base c prove that x+y+z=xyz-2

Answer 1

Let

$$x=\log_a(bc), \; y=\log_b(ac),\: z=\log_c(ab)$$

We put

$$\log_a(b)=u,\:\log_a(c)=v, \; \log_b(c)=w$$

Thus

$$x*y*z=[u+v]*[1/u+w]*[1/v+1/w]$$

$$=[1+v/u+uw+vw]*[1/v+1/w]$$

$$=1/v+1/u+uw/v+w+1/w+v/(uw)+u+v$$

Since

$$\log_a(b)*\log_b(c)=\log_a(c)$$

we get

$$v/(uw)=1$$

So

$$x*y*z=1/v+v+1/u+u+1/w+w+2$$

Also we have

$$x+y+z=[u+v]+[1/u+w]+[1/v+1/w]=1/v+v+1/u+u+1/w+w$$

Finally

$$x+y+z=x*y*z-2$$