Question

URGENTE MATEMATICA
Preciso que tenham certeza do que estao respondendo, é minha recuperação de matematica e não posso ir mal então por favor só responda quem realmente souber
preciso de todas as contas

Answer 1

Explicação passo-a-passo:

$a)\frac{1}{3^x} =\sqrt[4]{27} \\\\\frac{1}{3^x}=(3^3)^\frac{1}{4} \\\\3^-^x=(3)^\frac{3}{4} \\\\\\x=-\frac{3}{4}$

$b)\sqrt[3]{4^x} =\frac{1}{8} \\\\\sqrt[3]{2^2^x} =2^-^3 \\\\(2^\frac{2x}{3}) =2^-^3 \\\\\frac{2x}{3} =-3\\\\x=-\frac{9}{2}$

$c)\sqrt[3]{4^x^+^1} =1 \\\\(2^\frac{2x+2}{3}) =2^0 \\\\\ \frac{2x+2}{3} =0\\\\2x+2=0\\2x=-2\\x=-1$

$d)(0,16)^x=\sqrt[3]{\frac{25}{4} } \\\\(\frac{16}{100} )^x=(\frac{25}{4})^\frac{1}{3} \\\\\\(\frac{8}{50} )^x=(\frac{25}{4})^\frac{1}{3} \\\\(\frac{4}{25} )^x=(\frac{25}{4})^\frac{1}{3} \\\\(\frac{4}{25} )^x=(\frac{4}{25})^\frac{-1}{3} \\\\x=-\frac{1}{3}$

$e)(\sqrt{0,1} )^x=\sqrt[3]{10} \\(\sqrt{\frac{1}{10} } )^x=10^\frac{1}{3} \\\\10^-^\frac{x}{2} =10^\frac{1}{3} \\\\x=-\frac{2}{3}$

$f)(\frac{3}{5} )^-^x^+^2=\sqrt[3]{\frac{5}{3} } \\\\(\frac{3}{5} )^-^x^+^2=(\frac{5}{3})^\frac{1}{3} \\-x+2=\frac{1}{3} \\\\-x=\frac{1}{3} -2\\\\-x=\frac{1-6}{3} \\\\-x=\frac{-5}{3} \\\\x=\frac{5}{3}$

$g)(\frac{5}{3})^-^x^+^1 = \sqrt[3]{\frac{9}{25} } \\\\(\frac{5}{3})^-^x^+^1 = (\frac{5^2}{3^2} })^-^\frac{1}{3} \\\\(\frac{5}{3})^-^x^+^1 = (\frac{5}{3} })^-^\frac{2}{3} \\\\-x+1=\frac{-2}{3} \\-x=-\frac{2}{3}-1\\-x=\frac{-2-3}{3} \\-x=\frac{-5}{3} \\x=\frac{5}{3}$

$h)(\frac{1}{4})^5^-^x =\sqrt[3]{2^x} \\\\(2^-^2})^5^-^x =2^\frac{x}{3} \\\\-2.(5-x)=\frac{x}{3} \\-10+2x=\frac{x}{3} \\-30+6x=x\\6x-x=30\\5x=30\\x=\frac{30}{5} \\\\x=6$

$i)8^2^x^+^1=\sqrt[3]{4^x^-^1} \\2^3^(^2^x^+^1^)=2^\frac{2x-2}{3} \\6x+3=\frac{2x-2}{3} \\18x+9=2x-2\\18x-2x=-2-9\\16x=-11\\x=-\frac{11}{16}$

$j)9^2^x^+^1=\frac{1}{9^x} \\\\9^2^x^+^1=9^-^x\\2x+1=-x\\2x+x=-1\\3x=-1\\x=-\frac{1}{3}$

$l)5\sqrt[3]{5^2^x} =\frac{1}{125} \\\\5^1.5^\frac{2x}{3} =5^-^3\\\\5^\frac{2x}{3}^+^1 =5^-^3\\\frac{2x}{3} +1=-3\\\\\frac{2x}{3} =-3-1\\\\2x =-4.3\\\ x=-\frac{12}{2} \\\\\x=-6$

$m)\sqrt[\frac{5}{x} ]{32} =2\\\\(2^5)^\frac{x}{5} =2\\\\(2)^\frac{5x}{5} =2\\\\(2)^x =2\\\\x=1$