$${{log}}_{{\mathtt{0.5}}}{\left({\mathtt{19}}\right)}{\mathtt{\,\small\textbf+\,}}{{log}}_{{\mathtt{2}}}{\left({\mathtt{19}}\right)} = {\mathtt{0}}$$
how to explain this?
+128460 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) + log0.5 (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
+118608 Mmm interesting.
Let
$$\\x=log_{0.5}(19)\qquad and \qquad y=log_2(19)\\
so\\
19=0.5^x\qquad\qquad and \qquad 19=2^y\\
19=(\frac{1}{2})^x\\
2^x=\frac{1}{19}\\
so\\
2^x\times 2^y=\frac{1}{19}\times 19\\
2^x\times 2^y=1\\
2^x\times 2^y=2^0\\
2^{x+y}=2^0\\
$therefore$\\
x+y=0\\
so\\
log_{0.5}(19)+log_2(19)=0$$
+128460 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) + log0.5 (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
