What is the answer to 2^2x-3=(4^x)^2-3x-1 ?

The answer to 2^2x-3=(4^x)^2-3x-1 is No Solution for x\in\mathbb{R}

Solve for x: 2^2x-3=(4^x)^2-3x-1

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

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2^{2} x - 3 = (4^{x})^{2} - 3 x - 1

No Solution for x \in R

Solution steps

2^{2} x - 3 = (4^{x})^{2} - 3 x - 1

Prepare

2^{2} x - 3 = (4^{x})^{2} - 3 x - 1

for Lambert form:

1 = (7 x - 2) e^{- 4 l n (2) x}

Rewrite the equation with

4 ln (2) (- x + \frac{2}{7}) = u

and

x = - \frac{7 u - 8 ln ( 2 )}{28 ln ( 2 )}

1 = (7 (- \frac{7 u - 8 ln ( 2 )}{28 ln ( 2 )}) - 2) e^{- 4 l n (2) (- \frac{7 u - 8 l n ( 2 )}{28 l n ( 2 )})}

Simplify

1 = e^{\frac{7 u - 8 l n ( 2 )}{7}} (- \frac{7 u - 8 ln ( 2 )}{4 ln ( 2 )} - 2)

Rewrite

1 = e^{\frac{7 u - 8 l n ( 2 )}{7}} (- \frac{7 u - 8 ln ( 2 )}{4 ln ( 2 )} - 2)

in Lambert form:

e^{u} u = - \frac{4 e ^{\frac{8 l n ( 2 )}{7}} ln ( 2 )}{7}

Solve

e^{u} u = - \frac{4 e ^{\frac{8 l n ( 2 )}{7}} ln ( 2 )}{7} :

No Solution for

u \in R

No Solution for x \in R

2^{x} + 2^{3 x} = 16

2 e^{- x} = e^{x} - 1

x^{x} = 3

5^{x} = 3

2 5^{x - 1} = 5^{2 x - 1} - 100

What is the answer to 2^2x-3=(4^x)^2-3x-1 ?
The answer to 2^2x-3=(4^x)^2-3x-1 is No Solution for x\in\mathbb{R}
Solve for x: 2^2x-3=(4^x)^2-3x-1
The solution is No Solution for x\in\mathbb{R}