Question

ALGUÉM ME AJUDA A RESOLVER?

Answer 1

a) 3ˣ ⁺ ¹ = 27 => 3ˣ.3 = 27 => 3ˣ = 27/3 => 3ˣ = 9 => 3ˣ = 3² => x = 2

b) (0,25)²ˣ + ³ = 4ˣ ⁻ ¹ => (1/4)²ˣ ⁺ ³ = 4ˣ/4 => (4⁻¹)²ˣ ⁺ ³ = 4ˣ/4 => (4⁻¹)²ˣ.(4⁻¹)³ = 4ˣ/4 => 4⁻²ˣ.4⁻³ = 4ˣ/4 => 4⁻²ˣ/4³ = 4ˣ/4 => 4⁻²ˣ = 64.4ˣ/4 => 4⁻²ˣ = 16.4ˣ => 4⁻²ˣ - 16.4ˣ = 0 => (4ˣ)⁻² - 16.4ˣ = 0, fazendo 4ˣ = y temos

y⁻² - 16y = 0 => 1/y² - 16y =0 => 1 - 16y³ = 0 => -16y³ = -1, multiplicando ambos por -1, temos, 16y³ = 1 => y³ = 1/16 => y = ∛1/16 => y = 4$^{-\frac{2}{3} }$. Logo, 4ˣ = 4$-\frac{2}{3}$ => x = -2/3

c) √7ˣ = 343 => 7$^{\frac{x}{2} }$ = 7³ => x/2 = 3 => x = 6

d) 5$^{\frac{3}{2}x }$ = 125⁴ => 5$^{\frac{3}{2}x }$ = (5³)⁴ => 5$^{\frac{3}{2}x }$ = 5¹² => $\frac{3}{2}x$ = 12 => x = 24

e) $\sqrt[6]{1000}.10^{x} = 0,001 => \sqrt[6]{10^{3} }.10^{x} = 10^{-3} => \sqrt{10}.10^{x} = 10^{-3} => 10^{\frac{1}{2} }.10^{x} = 10^{-3} => 10^{\frac{1}{2} + x} = 10^{-3} => \frac{1}{2} + x = -3 => x = -3 - \frac{1}{2} => x = -\frac{7}{2}$

f) $\frac{8^{x} }{2^{x + 2} } = 256 => \frac{(2^{3})^{x}  }{2^{x + 2} } = 2^{8} => 2^{3x -x - 2} = 2^{8} => 2x - 2 = 8 => 2x = 10 => x = 10/2 => x = 5$