Question

alguém pode me ajudar a resolver essas três questões na moral?​

Answer 1

Resposta:

09)

a)

$\sf 5^{-x+1} = 1\\ 5^{-x+1} = 5^0\\-x + 1 = 0\\1 = x \\\framebox {x = 1}$

b)

$\sf 3^{2x + 7} = - 9 \\3^{2x + 7} = - 3^2 \\ 2x + 7 = - 2\\2x = - 2 - 7\\2x = - 9\\\\\boxed { \sf x = - \dfrac{9}{2}}$

10)

a)

$\sf \left ( \dfrac{3}{5} \right )^x = \left ( \dfrac{5}{3} \right )$

$\sf \left ( \dfrac{3}{5} \right )^x = \left ( \dfrac{3}{5} \right )^{-1}$

$\framebox { \sf x = - 1 }$

b)

$\sf 27^{x + 1} = 9^{- x + 5}\\3^3^{(x+ 1)} = 3^2^{(- x + 5)} \\3(x+ 1) = 2(- x + 5)\\3x + 3 = -2x + 10 \\3x + 2x = 10 - 3\\5x = 7 \\\\\boxed {\sf x = \dfrac{7}{5} }$

c)

$\sf \left ( \dfrac{2}{7} \right )^x = \dfrac{49}{4}$

$\sf \left ( \dfrac{2}{7} \right )^x =\left ( \dfrac{4}{49} \right)^{-1}$

$\sf \left ( \dfrac{2}{7} \right )^x =\left ( \dfrac{2}{7} \right)^{2^{(-1)}}$

$\sf x = 2(-1) \\\\\framebox { \sf x = - 2}$

d)

$\sf 2^{x^{2} - 4} = 0$ não solução  para x $\sf \in$ R.

11)

a)

$\sf (2^x)^2 = 16 \\2^{2x} = 2^4 \\2x = 4\\\\x= \dfrac{4}{2} \\\\\framebox {\sf x = 2 }$

b)

$\sf 2^{x^2} = 16 \\2^{x^2} = 2^4\\ x^{2} = 4 \\x = \pm \sqrt{4} \\\\\framebox { \sf x = - 4} \\\\\framebox { \sf x = 4 }$

c)

$\sf (3^x)^{2} = 3^{x^2} \\2x = x^{2} \\x^{2} = 2x \\\sf x^{2} -2x = 0\\\sf x(x-2) = 0\\\\\boxed{\sf x_1 = 0 }\\\\\sf (x - 2) = 0\\\sf x - 2 = 0\\\\\boxed {\sf x_2 = 2 }$

Explicação passo-a-passo: