Question

URGENTE ❌❌❌
sabendo que log 2= 0,301 e log 3= 0, 477 calcule:

a) log 16000
b) log 4,8
c) log 3√20​

Answer 1

Propriedades de logaritmo para essa questão:

• $\log a=$ log de a na base 10

• $\log_aa=1$

• $\log_a(b.c)=\log_ab+\log_ac$

• $\log_a(\frac{b}{c})=\log_ab-log_ac$

• $\log_ab^c=c.\log_ab$

• $a^\frac{b}{c}=\sqrt[c]{a^b}$

$\log16000=\log(2.2.2.2.1000)\\\\\log(2.2.2.2.1000)=\log2+\log2+\log2+\log2+\log1000\\\\\log2+\log2+\log2+\log2+\log10^3\\\\log2+\log2+\log2+\log2+3\log10\\\\0,301+0,301+0,301+0,301+3 = 4,204$

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$\log4,8=\log\frac{2.2.2.2.3}{10}\\\\ \log\frac{2.2.2.2.3}{10}=\log2+\log2+\log2+\log2+\log3-\log{10}\\\\0,301+0,301+0,301+0,301+0,477-1=0,681$

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$\log3\sqrt20=\log3+\log\sqrt20\\\\\log3+\log\sqrt20=\log3+\log20^\frac{1}{2}\\\\\log3+\log20^\frac{1}{2}= \log3+\log(10.2)^\frac{1}{2}\\\\\log3+\log(10.2)^\frac{1}{2}= \log3+\log10^\frac{1}{2}+\log2^\frac{1}{2}\\\\\log3+\log10^\frac{1}{2}+\log2^\frac{1}{2} = \log3+\frac{1}{2}\log10+\frac{1}{2}\log2\\\\0,477+\frac{1}{2}+\frac{0,301}{2}=1,1275$