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((12)^x)^2+4x-3>= 9

Solución

((12)^{x})^{2} + 4 x - 3 \geq 9

Solución

x \geq - \frac{W _{0} ( 1492992 ln ( 12 ) )}{2 ln ( 12 )} + 3

+2

Notación de intervalos

[- \frac{W _{0} ( 1492992 ln ( 12 ) )}{2 ln ( 12 )} + 3, \infty)

Notación decimal

x \geq 0.46602 \dots

Pasos de solución

((12)^{x})^{2} + 4 x - 3 \geq 9

Aplicar las leyes de los exponentes:

(a^{b})^{c} = a^{b c}

1 2^{2 x} + 4 x - 3

Mover

4 x

al lado derecho

1 2^{2 x} - 3 \geq 9 - 4 x

Mover

3

al lado derecho

1 2^{2 x} \geq - 4 x + 12

Preparar

1 2^{2 x} \geq - 4 x + 12

para la forma de Lambert:

1 \geq e^{- 2 l n (12) x} (- 4 x + 12)

Re-escribir la ecuación con

2 ln (12) (- x + 3) = u

y

x = - \frac{u - 6 ln ( 12 )}{2 ln ( 12 )}

1 \geq e^{- 2 l n (12) (- \frac{u - 6 l n ( 12 )}{2 l n ( 12 )})} (- 4 (- \frac{u - 6 ln ( 12 )}{2 ln ( 12 )}) + 12)

Simplificar

1 \geq e^{u - 6 l n (12)} (\frac{2 ( u - 6 ln ( 12 ) )}{ln ( 12 )} + 12)

Reescribir

1 \geq e^{u - 6 l n (12)} (\frac{2 ( u - 6 ln ( 12 ) )}{ln ( 12 )} + 12)

en forma de Lambert:

e^{u} u \leq 1492992 ln (12)

Resolver

e^{u} u \leq 1492992 ln (12) : u \leq W_{0} (1492992 ln (12))

u \leq W_{0} (1492992 ln (12))

Sustituir hacia atrás la

2 ln (12) (- x + 3) = u,

resolver para

x

Resolver

2 ln (12) (- x + 3) \leq W_{0} (1492992 ln (12)) : x \geq - \frac{W _{0} ( 1492992 ln ( 12 ) )}{2 ln ( 12 )} + 3

x \geq - \frac{W _{0} ( 1492992 ln ( 12 ) )}{2 ln ( 12 )} + 3

Ejemplos populares

1 + e^{x} > 0

(0.3)^{2x^2-3x-6}<0.00243

(0.3)^{2 x^{2} - 3 x - 6} < 0.00243

4^{x} < 2^{x} + 1 + 3

5^{1-x}<5^{4-5x}

5^{1 - x} < 5^{4 - 5 x}

solvefor x,(3/2)^{1+r*l^2}>(2/3)^{3x-1}resolver para

x, (\frac{3}{2})^{1 + r \cdot l^{2}} > (\frac{2}{3})^{3 x - 1}