Question

Calcule log2 x+log4 x+log8 x+log16 x= -6,25

Answer 1

Definição de logaritmos:

$log_{b}(a)=c~~~~~~~a,b~\textgreater~0,~a\neq1\\\\\\\boxed{\boxed{log_{b}(a)=c~~~\leftrightarrow~~~b^{c}=a}}$

Propriedade usada:

$\boxed{\boxed{log_{(b^{n})}(a)=\dfrac{1}{n}log_{b}(a)}}$
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$log_{2}(x)+log_{4}(x)+log_{8}(x)+log_{16}(x)=-6,25\\\\log_{2}(x)+log_{(2^{2})}(x)+log_{(2^{3})}(x)+log_{(2^{4})}(x)=-\frac{625}{100}\\\\log_{2}(x)+(\frac{1}{2})log_{2}(x)+(\frac{1}{3})log_{2}(x)+(\frac{1}{4})log_{2}(x)=-\frac{25}{4}$

O m.m.c entre 2, 3 e 4 é 12. Multiplicando todos os membros da equação por 12:

$12log_{2}(x)+12(\frac{1}{2})log_{2}(x)+12(\frac{1}{3})log_{2}(x)+12(\frac{1}{4})log_{2}(x)=12(-\frac{25}{4})\\\\12log_{2}(x)+6log_{2}(x)+4log_{2}(x)+3log_{2}(x)=3(-25)\\\\(12+6+4+3)log_{2}(x)=3(-25)\\\\25\cdot log_{2}(x)=3(-25)~~~(\div25)\\\\log_{2}(x)=-3$

Aplicando a definição de logaritmo:

$x=2^{-3}\\\\\\x=\dfrac{1}{2^{3}}\\\\\\\boxed{\boxed{x=\dfrac{1}{8}}}$