3^x+2/(3^{x-1)}=5

Solución

3^{x} + \frac{2}{3 ^{x - 1}} = 5

x = 1, x = \frac{ln ( 2 )}{ln ( 3 )}

Notación decimal

x = 1, x = 0.63092 \dots

Pasos de solución

3^{x} + \frac{2}{3 ^{x - 1}} = 5

Aplicar las leyes de los exponentes

3^{x} + 2 (3^{x})^{- 1} \cdot 3^{1} = 5

Re escribir la ecuación con

3^{x} = u

u + 2 (u)^{- 1} \cdot 3^{1} = 5

Resolver

u + 2 u^{- 1} \cdot 3^{1} = 5 : u = 3, u = 2

u = 3, u = 2

Sustituir hacia atrás la

u = 3^{x},

resolver para

x

Resolver

3^{x} = 3 : x = 1

Resolver

3^{x} = 2 : x = \frac{ln ( 2 )}{ln ( 3 )}

x = 1, x = \frac{ln ( 2 )}{ln ( 3 )}

Verificar las soluciones:

x = 1

Verdadero

, x = \frac{ln ( 2 )}{ln ( 3 )}

Verdadero

Las soluciones son

x = 1, x = \frac{ln ( 2 )}{ln ( 3 )}

625 = 2 5^{x - 10}

0.125 = 128 (\frac{1}{2})^{n}

e^{7 x} = 7

3.25 e^{\frac{0.2}{x}} = 30.6

4^{x + 1} + 2^{2 x - 1} = 288