Question

 Resolva a equação  :
Log₂X+Log₄X=4

Answer 1

$\sqrt[n]{a^{x}}=a^{x/n}$

$log_{b}(a)+log_{b}(c)=log_{b}(a*c)$
$log_{(b^{n})}(a)=(1/n)*log_{b}(a)$
$log_{b}(a)=c <=> b^{c}=a$
______________________

$log_{2}(x)+log_{4}(x)=4$
$log_{2}(x)+log_{(2^{2})}(x)=4$
$log_{2}(x) + (1/2)*log_{2}(x)=4$

Colocando log2(x) em evidência:

$(1 + [1 / 2])*log_{2}(x)=4$
$(3/2)*log_{2}(x)=4$
$log_{2}(x)=4*2/3$
$log_{2}(x)=8/3$
$2^{8/3}=x$
$x = \sqrt[3]{2^{8}}$
$x = \sqrt[3]{2^{6}}* \sqrt[3]{2^{2}}$
$x = 2^{6/3}* \sqrt[3]{4}$
$x = 2^{2}* \sqrt[3]{4}$
$x = 4 \sqrt[3]{4}$

Answer 2

$\\\log_2x+\log_4x=4\\\\\log_2x+\frac{\log_2x}{\log_24}=4\\\\\log_2x+\frac{\log_2x}{\log_22^2}=4\\\\\log_2x+\frac{\log_2x}{2}=4\\\\2\cdot\log_2x+\log_2x=8\\\\3\cdot\log_2x=8\\\\\log_2x=\frac{8}{3}\\\\2^{\frac{8}{3}}=x\\\\\boxed{x=\sqrt[3]{2^8}}$

 Ou, sob a forma apresentada pelo(a) Niiya.