Question

Dados que log9 base 18=a, calcule o valor de log de 2 base 9 em função de a

Answer 1

$log_{18}(9)=a\\\\\\log_{18}\left(\dfrac{18}{2}\right)=a\\\\\\log_{18}(18)-log_{18}(2)=a\\\\\\1-log_{18}(2)=a\\\\\\\boxed{\boxed{log_{18}(2)=1-a}}$

Agora, vamos usar a propriedade $log_{x}(y)=\dfrac{1}{log_{y}(x)}$:

$\dfrac{1}{log_{2}(18)}=1-a~~~\therefore~~~log_{2}(18)=\dfrac{1}{1-a}\\\\\\log_{2}(2\cdot9)=\dfrac{1}{1-a}\\\\\\log_{2}(2)+log_{2}(9)=\dfrac{1}{1-a}\\\\\\1+log_{2}(9)=\dfrac{1}{1-a}\\\\\\log_{2}(9)=\dfrac{1}{1-a}-1\\\\\\log_{2}(9)=\dfrac{1-(1-a)}{1-a}\\\\\\log_{2}(9)=\dfrac{1-1+a}{1-a}\\\\\\log_{2}(9)=\dfrac{a}{1-a}$

Utilizando novamente a propriedade:

$\boxed{\boxed{log_{9}(2)=\dfrac{1-a}{a}=\dfrac{1}{a}-1}}$

Answer 2

Dado: 
log(18) 9 = a

Mudando para a base 9: 

log(9) 9/ log(9) 18 = a
1 / log(9) (2.9) = a
1 / [log(9) 2 + log(9) 9] = a
1 / [log(9) 2 + 1] = a
log(9) 2 + 1 = 1/a
log(9) 2 = 1/a - 1