Question

alguem ajuda preciso dos calculos

Answer 1

a)

$3^{x-2} = 27 \\ \\ 3^{x-2} = 3^3 \\ \\ x-2 = 3 \\ \\ x = 5$

b)

$4^{x+5} = 8 \\ \\ (2^2)^{x+5} = 2^3 \\ \\ 2^{2x + 10} = 2^3 \\ \\ 2x + 10 = 3 \\ \\ x = \frac{-7}{2}$

c)

$(5^{x+3})^{x-3} = 1 \\ \\ 5^{ x^{2} - 9 } = 5^0 \\ \\ x^2 - 9 = 0 \\ \\ x^2 = 9 \\ \\ x' = 3 \\ x'' = -3$

e)

$9^x - 4*3^x + 3 = 0 \\ \\ (3^{x})^2 - 4*3^x + 3 = 0 \\ \\ y = 3^x \\ \\ y^2 - 4y + 3 = 0$

Δ = 16 - (4*3)
Δ = 16 - 12
Δ = 4

$y' = \frac{4+2}{2} = 3 \\ \\ y'' = \frac{4-2}{2} = 1$

Como:

$y = 3^x \\ \\ 3 = 3^x' \\ x' = 1 \\ \\ 1=3^x'' \\ 3^0 = 3^{x''} \\ x'' = 0$

f)

$4^x - 5*2^x +4 = 0 \\ \\ (2^x)^{2} - 5*2^x + 4 = 0 \\ \\ y = 2^x \\ \\ y^2 - 5y + 4 = 0$

Δ = 25 - (4*4)
Δ = 25-16
Δ = 9

$y' = \frac{5 + 3}{2} = 4 \\ \\ y'' = \frac{5-3}{2} = 1$

Como:

$y =2^x \\ \\ 4 = 2^x' \\ 2^2 = 2^x' \\ x' = 2 \\ \\ 1 = 2^x'' \\ 2^0 = 2^x'' \\ x'' = 0$

g)

$4^x - 2^x - 2 = 0 \\ \\ (2^x)^2 - 2^x - 2 = 0 \\ \\ y = 2^x \\ \\ y^2 - y - 2 = 0$

Δ = 1 + (4*2)
Δ = 1 + 8
Δ = 9

$y' = \frac{1 + 3}{2} = 2 \\ \\ y'' = \frac{1-3}{2} = -1 \ \ \ \ (negativo \ descarta)$

Como:

$y = 2^x \\ \\ 2 = 2^x \\ x = 1$