log(19, 0.5) + log(19, 2) = 0 how to explain this?

Here's a slightly different approach to Melody's

Note that we can write .5  as 1/2  = 2^-1 

So we have

log₂(19)  + log_0.5 (19)   .....and...using the"change of base" rule, we can write

log(19) / log(2)  +  log(19) / log(.05) =

log(19) / log(2)  + log(19) / log(1/2) =

log(19) / log(2)  +  log(19) / log2^-1 =

log(19) / log(2)  +  log(19) / (-1) log2 =

log(19) / log(2)  +  [ (-1) log(19) /  log2 ] = 

log(19) / log(2)  -  log(19) /  log2 =   0

Mmm interesting.

Let 

$$\\x=log_{0.5}(19)\qquad and \qquad y=log_2(19)\\ so\\ 19=0.5^x\qquad\qquad and \qquad 19=2^y\\ 19=(\frac{1}{2})^x\\ 2^x=\frac{1}{19}\\ so\\ 2^x\times 2^y=\frac{1}{19}\times 19\\ 2^x\times 2^y=1\\ 2^x\times 2^y=2^0\\ 2^{x+y}=2^0\\ $therefore$\\ x+y=0\\ so\\ log_{0.5}(19)+log_2(19)=0$$

Thanks Chris 

It is always good to get a couple of different takes