Question

bom estou com dificuldades de responder e queria com graficos prfvr​

Answer 1

16 -

 Propriedades utilizadas:

$x^{\frac{a}{b}}=\sqrt[b]{x^a}\\x^{-a} = \frac{1}{x^a}\\(x^a)^b = x^{a.b}$

 Nosso objetivo será igualar as bases dos dois lados da equação, pois se as bases são iguais, os expoentes também terão que ser iguais para que a igualdade seja verdadeira.

a)

$(\sqrt{2})^{3.x-1} =(\sqrt[3]{16})^{2.x-1}\\(2^{\frac{1}{2}})^{3.x-1}=(\sqrt[3]{2^4})^{2.x-1}\\2^{\frac{3.x-1}{2}} = (2^{\frac{4}{3}})^{2.x-1}\\2^{\frac{3.x-1}{2}}=2^{\frac{4.(2.x-1)}{3}}\\2^{\frac{3.x-1}{2}}=2^{\frac{8.x-4}{3}}\\\\\frac{3.x-1}{2}=\frac{8.x-4}{3}\\3.(3.x-1)=2.(8.x-4)\\9.x-3 = 16.x-8\\9.x -16.x = -8 +3\\-7.x = -5\\x = \frac{5}{7}$

b)

$8^{2.x-1}=\sqrt{4^{x-1}} \\2^{3.(2.x-1)}=\sqrt{2^2.(x-1)}\\2^{6.x-3}=\sqrt{2^{2.x-2}}\\2^{6.x-3}=2^{\frac{2.x-2}{2}}\\\\6.x -3=\frac{2.x-2}{2}\\2.(6.x-3)=2.x-2\\12.x-6=2.x-2\\12.x-2.x = -2+6\\10.x = 4\\x = \frac{4}{10}\\x=\frac{2}{5}$

c)

$\frac{1}{7}=\sqrt{49^{x-1}} \\7^{-1}=\sqrt{7^{2.(x-1)}}\\7^{-1}=\sqrt{7^{2.x-2}}\\7^{-1}=7^{\frac{2.x-2}{2}}\\\\-1=\frac{2.x-2}{2}\\-2=2.x-2\\2.x = -2 +2\\2.x = 0\\x=0$

d)

$(3^x)^x=9^8\\3^{x.x}=3^{2.8}\\3^{x^2}=3^{16}\\\\x^2=16\\\\x' = 4\\x'' = -4$

17 -

$(0,0625)^{x+2}=0,25\\(\frac{625}{1000})^{x+2}=\frac{25}{100}\\(\frac{5^4}{10^4})^{x+2}=\frac{5^2}{10^2}\\(\frac{5}{10})^{4.(x+2)}=(\frac{5}{10})^2\\(\frac{5}{10})^{4.x+8}=(\frac{5}{10})^2\\\\4.x +8 = 2\\4.x = 2-8\\4.x = -6\\x = \frac{-6}{4}\\x=-\frac{3}{2}\\\\(-\frac{3}{2}+1)^6\\(\frac{-1}{2})^6\\\frac{1}{64}$

18 -

$\sqrt[3]{\frac{2^{28}+2^{30}}{10}} \\\sqrt[3]{\frac{2^{28}.(1+2^2)}{10}}\\\sqrt[3]{\frac{2^{28}.(1+4)}{10}} \\\sqrt[3]{\frac{2^{28}.5}{10}} \\\sqrt[3]{\frac{2^{28}}{2}} \\\sqrt[3]{2^{28-1}} \\\sqrt[3]{2^{27}}\\2^{\frac{27}{3}} \\2^9\\512$

19 -

$\frac{21^{30}}{63^{15}}\\\\\frac{(3.7)^{30}}{(9.7)^{15}}\\\\\frac{3^{30}.7^{30}}{9^{15}.7^{15}}\\\\\frac{3^{30}.7^{30}}{3^{2.15}.7^{15}}\\\\\frac{3^{30}.7^{30}}{3^{30}.7^{15}}\\\\\frac{7^{30}}{7^{15}}\\\\7^{30-15}\\7^{15}$

20 -

$a^{\frac{1}{2}}+a^{-\frac{1}{2}}= \frac{10}{3}\\\sqrt{a}+\frac{1}{\sqrt{a}}=\frac{10}{3}\\\frac{a+1}{\sqrt{a}}=\frac{10}{3}\\3.(a+1)=10.\sqrt{a}\\3.a+3=10.\sqrt{a}\\\frac{3.a+3}{10}=\sqrt{a}\\(\frac{3.a+3}{10})^2=(\sqrt{a})^2\\\frac{9.a^2+18.a+9}{100}=a\\9.a^2+18.a+9=100.a\\9.a^2-82.a+9=0\\\\\\D=b^2-4.a.cD=(-82)^2-4.9.9\\D=6724-4.81\\D=6724-324\\D =6400\\\\a=\frac{-b+-\sqrt{D}}{2.a}\\a=\frac{82+-\sqrt{6400}}{2.9}\\a=\frac{82+-80}{18}$

$a'=\frac{82-80}{18}\\a'=\frac{2}{18}\\a'=\frac{1}{9}\\\\a''=\frac{82+80}{18}\\a''\frac{162}{18}\\a''=9\\\\a+a^{-1}\\\frac{1}{9}+\frac{1}{\frac{1}{9}}\\\frac{1}{9}+9\\\frac{82}{9}\\\\a+a^{-1}\\9+\frac{1}{9}\\\frac{82}{9}$

Dúvidas só perguntar XD