Question

Sendo a e b dois numeros reais tais que log2 (log3 a) = 1 . Log3 (log2 b) = 0, temos a+b igual a:

Gabarito é 11! Preciso da resolução

Answer 1

Boa noite Diego!

Solução!

Vamos encontrar o valor de a e b,usando as propriedades dos logaritmos.
Veja que teremos que escrever os logaritmos na base 10 usando a mudança de base.


$log2(log_{3}a)=1\\\\\\log2. \bigg(\dfrac{loga}{log3}\bigg)=1 \\\\\\\\ log2+loga-log3=1\\\\\\ log(2+a-3)=1\\\\\\ 2+a-3=10^{1} \\\\\\ 2+a-3=10\\\\\ a-3+2=10\\\\\ a-1=10\\\\\\ a=10+1\\\\\ \boxed{a=11}$



$log3(log_{2}b)=0\\\\\\\\ log3.\bigg( \dfrac{logb}{log2}\bigg)=0\\\\\\\ log3+logab-log2=0\\\\\\ log(3+b-2)=0\\\\\\ (3+b-2)=10^{0} \\\\\\\ b+3-2=1\\\\\\ b+1=1\\\\\\ b=1-1\\\\\\ \boxed{b=0}$


$~~a~~+~~b\\\\\ \boxed{11~~+~0=11}$

Boa noite!
Bons estudos!