Question

Resolva a equação logarítmica no campo dos números reais,

$log_{16} \sqrt{x} + log_4\sqrt[4]{x}+ \dfrac{1}{4} log_{0,5}x=log_2x$

Answer 1

$log_{16}\sqrt{x}+log_{4}\sqrt[4]{x}+\dfrac{1}{4}log_{0.5}x=log_{2}x\\\\\\log_{(2^{4})}(x^{1/2})+log_{(2^{2})}(x^{1/4})+\dfrac{1}{4}log_{(2^{-1})}x=log_{2}x$

Sabemos que:

$\boxed{log_{(b^{n})}(a)=\frac{1}{n}*log_{b}(a)}~~\boxed{log_{b}(a^{n})=n*log_{b}(a)}$

Logo:

$\dfrac{1}{4}\cdot\dfrac{1}{2}\cdot log_{2}x+\dfrac{1}{2}\cdot\dfrac{1}{4}\cdot log_{2}x+\dfrac{1}{4}\cdot\dfrac{1}{-1}\cdot log_{2}x=log_{2}x\\\\\\2\cdot\dfrac{1}{8}log_{2}x-\dfrac{1}{4}log_{2}x=log_{2}x\\\\\\\dfrac{1}{4}log_{2}x-\dfrac{1}{4}log_{2}x=log_{2}x\\\\\\log_{2}x=0$

Pela definição de logaritmos:

$2^{0}=x\\\\\boxed{\boxed{x=1}}$

Answer 2

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$\sf{\ell og_{16}\sqrt{x}+\ell og_4\sqrt[4]{x}+\dfrac{1}{4}\ell og_{0,5}x=\ell og_2x}\\\sf{\dfrac{1}{4}\cdot\dfrac{1}{2}\ell og_2x+\dfrac{1}{2}\cdot\dfrac{1}{4}\ell og_2x+\dfrac{1}{4}\cdot(-1)\ell og_2x=\ell og_2x\cdot(8)}\\\sf{\ell og_2x+\ell og_2x-2\ell og_2x=8\ell og_2x}\\\sf{8\ell og_2x=\diagdown\!\!\!\!\!\!\!\!2\ell og_2x-\diagdown\!\!\!\!\!\!\!\!2\ell og_2x}\\\sf{8\ell og_2x=0}\\\sf{\ell og_2x=\dfrac{0}{8}}\\\sf{\ell og_2x=0}\\\sf{x=2^0}\\\Large\boxed{\boxed{\boxed{\boxed{\boxed{\sf{x=1}}}}}}$