Question

Se o Log60 3=x que o Log60 6=y, qual é o Log18 2

Answer 1

$\bmatrix log_{60}(3)=x\\\\log_{60}(6)=y\end$
 

como 6 = 3*2 
então

$log_{60}(6)=y\\\\log_{60}(3*2)=y\\\\log_{60}(3)+log_{60}(2)=y\\\\\boxed{\log_{60}(2)=y-x}\\\\.$

resolvendo
$log_{18}(2)$

mudando pra base 60

$log_{18}(2)= \frac{log_{60}(2)}{log_{60}(18)} =\\\\= \frac{y-x}{log_{60}(18)} = \frac{y-x}{log_{60}(9*2)} = \frac{y-x}{log_{60}(9)+log_ {60(2) }} =\\\\= \frac{y-x}{log_{60}(3^2)+(y-x)} = \frac{y-x}{2*log_{60}(3)+y-x} = \frac{y-x}{2x+y-x} = \frac{y-x}{x+y} \\\\.$

Answer 2

Faça uma mudança de base:

log18(2)=log60(2)/log60(18)
=log60(2)/log60(3.6)
=log60(2)/log60(3)+log60(6)
=log60(2)/x+y
=log60(6/3)/x+y
=log60(6)-log60(3)/x+y
=y-x/x+y

logo:

log18(2)=y-x/x+y