What is the answer to 2^{x+2}-3x(2^{x-1})=20 ?

The answer to 2^{x+2}-3x(2^{x-1})=20 is No Solution for x\in\mathbb{R}

Solve for x: 2^{x+2}-3x(2^{x-1})=20

The solution is No Solution for x\in\mathbb{R}

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2^{x + 2} - 3 x (2^{x - 1}) = 20

No Solution for x \in R

Solution steps

2^{x + 2} - 3 x (2^{x - 1}) = 20

Prepare

2^{x + 2} - 3 x (2^{x - 1}) = 20

for Lambert form:

e^{l n (2) (- 1 + x)} (8 - 3 x) = 20

Rewrite the equation with

(x - \frac{8}{3}) ln (2) = u

and

x = \frac{u}{ln ( 2 )} + \frac{8}{3}

e^{l n (2) (- 1 + (\frac{u}{l n ( 2 )} + \frac{8}{3}))} (8 - 3 (\frac{u}{ln ( 2 )} + \frac{8}{3})) = 20

Simplify

- \frac{3 e ^{\frac{5}{3} l n (2) + u} u}{ln ( 2 )} = 20

Rewrite

- \frac{3 e ^{\frac{5}{3} l n (2) + u} u}{ln ( 2 )} = 20

in Lambert form:

\frac{e ^{\frac{3 u + 5 l n ( 2 )}{3}} u}{e ^{\frac{5 l n ( 2 )}{3}}} = - \frac{20 ln ( 2 )}{3 e ^{\frac{5 l n ( 2 )}{3}}}

Solve

\frac{e ^{\frac{3 u + 5 l n ( 2 )}{3}} u}{e ^{\frac{5 l n ( 2 )}{3}}} = - \frac{20 ln ( 2 )}{3 e ^{\frac{5 l n ( 2 )}{3}}} :

No Solution for

u \in R

No Solution for x \in R

(1 + x) \cdot 4 = 2^{x}

2187 = \frac{3 ^{n} - 1}{2}

10 = 95 \cdot 0.91 4^{x}

5^{t} = 7

What is the answer to 2^{x+2}-3x(2^{x-1})=20 ?
The answer to 2^{x+2}-3x(2^{x-1})=20 is No Solution for x\in\mathbb{R}
Solve for x: 2^{x+2}-3x(2^{x-1})=20
The solution is No Solution for x\in\mathbb{R}