Question

resolver las siguentes ecuaciones exponenciales​

Answer 1

Respuesta:

Pregunta 5

${2}^{3 {x}^{2} + 11x } = 16 \\ {2}^{3 {x}^{2} + 11x } = {2}^{4} \\ 3 {x}^{2} + 11x = 4 \\ 3 {x}^{2} + 11x - 4 = 0$

∆ = (11)² - 4(-4)(3)

∆ = 121 + 48

∆ = 169

$x = \frac{ - 11 \binom{ + }{ - } \sqrt{169} }{6} \\ x = \frac{ - 11 \binom{ + }{ - } 13 }{6}$

X(1) = (-11 + 13)/6 = 2/6 = 1/3

X(2) = (-11 - 13)/6 = -24/6 = -4

• Como x no puede ser negativo la respuesta es 1/3

Pregunta 6

${( \frac{1}{2} )}^{4x - 2} = {2}^{x - 1} \\(( {2} ^{-1}) ^{4x - 2}) = {2}^{x - 1} \\ {2}^{-4x+2} = {2}^{x - 1} \\ - 4x + 2 = x - 1 \\ 2 + 1 = x + 4x \\ 3 = 5x \\ \frac{3}{5} = x$

Pregunta 7

${9}^{ \sqrt{2x - 1} } - {3}^{ \sqrt{2x - 1} + 3 } = 0 \\ {9}^{ \sqrt{2x - 1} } = {3}^{ \sqrt{2x - 1} + 3 } \\ {( {3}^{2} )}^{ \sqrt{2x - 1} } ={3}^{ \sqrt{2x - 1} + 3 } \\ {3}^{2 \sqrt{2x - 1} } = {3}^{ \sqrt{2x - 1} + 3 } \\ 2 \sqrt{2x - 1} = \sqrt{2x - 1} + 3 \\ 2 \sqrt{2x - 1} - \sqrt{2x - 1} = 3 \\ \sqrt{2x - 1} = 3 \\ 2x - 1 = {3}^{2} \\ 2x = 9 + 1 \\ x = \frac{10}{2} \\ x = 5$

Pregunta 8

${4}^{2x} - {2}^{2x} - 12 = 0 \\ ({2}^{2})^{2x} - {2}^{2x} - 12 = 0 \\ {2}^{2x} = a \\ {a}^{2} - a - 12 = 0$

a² - a - 12 = 0

a ..........-4 = - 4a

a ........... 3 = 3a

(a - 4)(a + 3) = 0

a = 4 y a = -3

2^2x = 4

2^2x = 2²

2x = 2

x = 2/2

x = 1