Question

Me ajudem por favor?
a número 1 eu só preciso da letra E).​

Answer 1

Resposta:

1 e) x = -3/4

2 a) x = 9

b ) x = -13

c) x = 0

d) x = 1

e) x = 0 ou x = 1

f) x = 0

3) a) x = 1 ou x = 2

b) x = 1/5

c) x = 0 ou x = 3

d) x = 2

e) x = 2 ou x = 0

Explicação passo-a-passo:

1 e)

343/7 = 49

entao 343 = 49 x 7 = 49^(2x+3)

mas 49 = 7^2 entao

$7^2.7 = (7^2)^{(2x+3)}\\\\7^3 = 7^{2(2x+3)}\\\\7^3 = 7^{4x+6}\\\\3 = 4x+6\\\\4x = -3\\\\x = -\frac{3}{4}$

2 a) 125^(x-9) = 1

qualquer numero elevado a 0 = 1, entao

x - 9 = 0

x = 9

b) 2^(x+7) = 1/64

64 = 8^2 = (2^3)^2 = 2^6

entao 1/64 = 2^(-6)

logo

x+7 = -6

x = -13

c) lembra que a^m • a^n = a^(m+n)? vamos usar isso aqui...

$3^{x+4} + 3^{x+2} = 90\\\\3^x 3^4 + 3^x 3^2 = 90\\\\3^x (3^4 + 3^2) = 90\\\\3^x (81 + 9) = 90\\\\3^x = 90/90\\\\3^x = 1\\\\x = 0$

d)

$5^{x-1} + 5^{x+1} = 26\\\\5^x 5^{-1} + 5^x 5^1 = 26\\\\5^x (5^{-1} + 5^1) = 26\\\\5^x (1/5 + 5) = 26\\\\5^x (26/5) = 26\\\\5^x = 5\\\\x = 1$

e)

$10^{2x-1} - 11.10^{x-1} + 1 = 0\\\\10^{-1} = 1/10\\\\\frac{10^{2x}}{10} - 11 \frac{10^x}{10} = -1\\\\10^{2x} - 11.10^x = -10\\\\a^{nx} = (a^x)^n\\\\(10^x)^2 - 11.10^x+10 = 0\\\\10^x = a\\\\a^2 - 11a + 10 = 0\\\\delta = (-11)^2 - 4(1)(10) = 121 - 40 = 81\\\\a' = \frac{11 +\sqrt{81} }{2}\\\\a' = \frac{11+9}{2}\\\\a' = 10\\\\a" = \frac{11 - 9}{2} \\\\a" = 1$

entao,

$10^x = 1\\\\x = 0\\\\10^x = 10\\\\x = 1$

f)

$5^{2x} + 4.5^x - 5 = 0\\\\5^x = a\\\\a^2 + 4a - 5 = 0\\\\a' = \frac{-4 + \sqrt{16 - (4.1.-5)} }{2} \\\\a' = \frac{-4+\sqrt{16 + 20} }{2} \\\\a' = \frac{-4+6}{2} \\\\a'= 1\\\\a" = -5\\\\5^x = 1\\\\x = 0\\\\5^x = -5$

-5 nao é solucao... portanto x = 0

3)

$4^x = 5.2^x -4\\\\(2^2)^x = 5.2^x - 4\\\\2^{2x} - 5.2^x + 4 = 0\\\\2^x = a\\\\a^2 - 5a + 4 = 0\\\\a'= \frac{5+\sqrt{25 - 16} }{2}\\\\a' = \frac{5+3}{2} \\\\a'=4\\\\a" = 1$

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

b)

$5^{10x} - 10.5^{5x} + 25 = 0\\\\5^{ 5x} = a\\\\a^2 - 10a + 25 = 0\\\\a' = \frac{10+\sqrt{100-100} }{2} \\\\a' = a" = 5\\\\5^5x = 5^1\\\\5x = 1\\\\x = \frac{1}{5}$

c)

$4^x - 9.2^x = -8\\\\2^x = a\\\\a^2 -9a+8 = 0\\\\a'= \frac{9+\sqrt{81-32} }{2}\\\\a'= \frac{9+\sqrt{49} }{2} \\\\a'= 8\\\\a" = 1\\\\2^x = 8\\x = 3\\\\2^x = 1\\\\x = 0$

d)

$16^x - 48 = 13.4^x\\\\4^x = a\\\\a^2 - 13a - 48 = 0\\\\a' = \frac{13+\sqrt{169+192} }{2} \\\\a' = \frac{13+\sqrt{361} }{2}\\\\a'=\frac{13+19 }{2} \\\\a'= 16\\\\a"= -3\\\\4^x = 16\\\\x = 2$

e)

$3^x = a\\\\a^2 - 10a + 9 = 0\\\\a'= \frac{10+\sqrt{100-36} }{2}\\\\a'= \frac{10+ 8}{2} \\\\a' =9\\\\a"= 1\\\\3^x = 9\\\\x = 2\\\\3^x = 1\\\\x = 0$