Question

Aplicando as definições, calcule o valor dos logaritmos:
a)
$log \sqrt {8}^{4}$
b)
$log_{5}0.000064$
c)
$log_{49} \sqrt[3]{7}$
d)
$log_{2} \frac{1}{16}$
e)
$log\frac{1}{2} \sqrt[3]{32}$
f)
$log_ \frac{1}{4} 1024$
ME AJUDEM PLMDS.​

Answer 1

Resposta:

a)

$\log_\sqrt{8} }4=x\\ \\ (\sqrt{8} )^x=4\\ \\ (8^{{1\over2}})^x=4\\ \\ (2^{3.{1\over2}})^x=4\\ \\ (2^{3\over2})^x=2^2\\ \\ \not2^{3x\over2}=\not2^2\\ \\ \dfrac{3x}{2}=2\\ \\ 3x=4\\ \\ x=\dfrac{4}{3}\\ \\ \\ \log_\sqrt{8} }4=\dfrac{4}{3}}}}$

b)

$\log_50,000064=x\\ \\ 5^x=0,000064\\ \\ 5^x=64.10^{-6}\\ \\ 5^x=2^6.2^{-6}.5^{-6}\\ \\ \not5^x=\not5^{-6}\\ \\ x=-6$

c)

$\log_{49}\sqrt[3]{7} =x\\ \\ 49^x=\sqrt[3]{7} \\ \\ \not7^{2x}=\not7^{1\over3}\\ \\ 2x=\dfrac{1}{3}\\ \\ 6x=1\\ \\ x=\dfrac{1}{6}$

d)

$\log_2{1\over16}=x\\ \\ 2^x=\dfrac{1}{2^4}\\ \\ \not2^x=\not2^{-4}\\ \\ x=-4$

e)

$\log_{1\over2}\sqrt[3]{32} =x\\ \\ ({1\over2})^x=\sqrt[3]{2^5} \\ \\ \not2^{{-x}}=\not2^{5\over3}\\ \\ -x=\dfrac{5}{3}\\ \\ x=-\dfrac{5}{3}$

f)

$\log_{1\over4}1024=x\\ \\ ({1\over4})^x=1024\\ \\ ({1\over2^2})^x=2^{10}\\ \\ \not2^{-2x}=\not2^{10}\\ \\ -2x=10\\ \\ 2x=-10\\ \\ x=-\dfrac{10}{2}\\ \\ x=-5$