Calcule o valor da soma S :a) log 0,001 + log 2^8 - log 3^1b) log 1/2^1/64 + log √3^81 + log 2√8.... Pergunta de ideia deJoonSC

March 2022 1 46 Report

Calcule o valor da soma S :
a) log 0,001 + log 2^8 - log 3^1
b) log 1/2^1/64 + log √3^81 + log 2√8

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korvo

Verified answer

E ae mano,

suponho que seja assim que quis escrever:

$S=log0,001+log_28-log_31\\
S=log_{10}0,001+log_22^3-log_33^0~~~~.\\\\
S=log_{10} \dfrac{1}{1.000}+log_22^3-log_33^0\\\\
S=log_{10}10^{-3}+log_22^3-log_33^0\\
S=-3\cdot(log_{10}10)+3\cdot log_22-0\cdot log_33\\
S=-3\cdot1+3\cdot1-0\cdot1\\
S=-3+3-0\\\\
\Large\boxed{\boxed{S=0}}|\\-$

>>>>>>>>>>>>>>>>>>

$S= log_{\tfrac{1}{2}} \dfrac{1}{64}+log_{ \sqrt{3}}81+log_2 \sqrt{8}\\\\
S=-log_22^{-4}+log_{ \sqrt{3}}3^4+log_2 \sqrt{2^3}\\\\
S=-4\cdot(-log_22)+4\cdot log_ {\sqrt{3}}3+log_22^{ \tfrac{3}{2}}\\\\
S=4\cdot log_22+4\cdot log_{3^{ \tfrac{1}{2}}}3+ \dfrac{3}{2}\cdot log_22\\\\
S=4\cdot log_22+4\cdot2\cdot log_33+ \dfrac{3}{2}\cdot log_22\\\\
S=4\cdot1+8\cdot1+ \dfrac{3}{2}\cdot1\\\\
S=4+8+ \dfrac{3}{2}\\\\
\Large\boxed{\boxed{S= \dfrac{27}{2}}}|\\-$

Tenha ótimos estudos mano ;D

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