Question

Determine o valor de x na expressão abaixo.

$x =\dfrac{\left(\frac{({8}^{-2}) \cdot\sqrt{16}}{{8}^{-2} }\right)\cdot{16}^{-\frac{1}{2}}\frac{\sqrt{64}}{{2}^{4}\cdot{2}^{-1}} }{\left ( log_{4}( log_{3}(9)) + log_{2}( log_{81}(3) ) + log_{0.8}( log_{16}(32) ) \right )\cdot log_{2}( \sqrt{2} ) }$
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Answer 1

8^-2×√16/8^-2=√16=4

16^-1/2 ×√64/2⁴×2^-1=1/√16 ×8/2³=1/4

log4(log3(9))=log4(2)=1/2

log2(log81(3))=log2(1/4)=-2

log0,8(log16(32))=log0,8(5/4)=-1

log2(√2)=1/2

x=(4×1/4)/((1/2-2-1)×1/2)

x=1/-5/4

x=-4/5

Answer 2

Explicação passo-a-passo:

$\sf x=\dfrac{\frac{8^{-2}\cdot\sqrt{16}}{8^{-2}}\cdot16^{\frac{-1}{2}}\cdot\frac{\sqrt{64}}{2^4\cdot2^{-1}}}{[log_{4}~(log_{3}~9)+log_{2}~(log_{81}~3)+log_{0,8}~(log_{16}~32)]\cdot log_{2}~\sqrt{2}}$

$\sf x=\dfrac{\frac{\not{8^{-2}}\cdot4}{\not{8^{-2}}}\cdot\frac{1}{4}\cdot\frac{8}{8}}{[log_{4}~2+log_{2}~\frac{1}{4}+log_{0,8}~\frac{5}{4}]\cdot\frac{1}{2}}$

$\sf x=\dfrac{4\cdot\frac{1}{4}\cdot1}{[\frac{1}{2}-2-1]\cdot\frac{1}{2}}$

$\sf x=\dfrac{1}{[\frac{1}{2}-2-1]\cdot\frac{1}{2}}$

$\sf x=\dfrac{1}{-\frac{5}{2}\cdot\frac{1}{2}}$

$\sf x=\dfrac{1}{-\frac{5}{4}}$

$\sf \red{x=-\dfrac{4}{5}}$

* $\sf log_{a^m}~a^n=\dfrac{m}{n}\cdot log_{a}~a=\dfrac{m}{n}$

• $\sf log_{3}~9=log_{3}~3^2=2$

• $\sf log_{81}~3=log_{3^4}~3^1=\dfrac{1}{4}$

• $\sf log_{16}~32=log_{2^4}~2^5=\dfrac{5}{4}$

• $\sf log_{2}~\sqrt{2}=log_{2}~2^{\frac{1}{2}}=\dfrac{1}{2}$

• $\sf log_{4}~2=log_{4}~4^{\frac{1}{2}}=\dfrac{1}{2}$

• $\sf log_{2}~\dfrac{1}{4}=log_{2}~2^{-2}=-2$

• $\sf log_{0,8}~\dfrac{5}{4}=log_{0,8}~0,8^{-1}=-1$