What is the answer to 2^2x-2^x+3+2^4=0 ?

The answer to 2^2x-2^x+3+2^4=0 is

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2^2x-2^x+3+2^4=0

2^{2} x - 2^{x} + 3 + 2^{4} = 0

x = - \frac{4 W _{- 1} ( - \frac{l n ( 2 )}{4 e ^{\frac{19 l n ( 2 )}{4}}} ) + 19 ln ( 2 )}{4 ln ( 2 )}, x = - \frac{4 W _{0} ( - \frac{l n ( 2 )}{4 e ^{\frac{19 l n ( 2 )}{4}}} ) + 19 ln ( 2 )}{4 ln ( 2 )}

Decimal Notation

x = 5.33399 \dots, x = - 4.74064 \dots

Solution steps

2^{2} x - 2^{x} + 3 + 2^{4} = 0

Prepare

2^{2} x - 2^{x} + 3 + 2^{4} = 0

for Lambert form:

- 1 = - e^{- l n (2) x} (4 x + 19)

Rewrite the equation with

(- x - \frac{19}{4}) ln (2) = u

and

x = - \frac{4 u + 19 ln ( 2 )}{4 ln ( 2 )}

- 1 = - e^{- l n (2) (- \frac{4 u + 19 l n ( 2 )}{4 l n ( 2 )})} (4 (- \frac{4 u + 19 ln ( 2 )}{4 ln ( 2 )}) + 19)

Simplify

- 1 = - e^{\frac{4 u + 19 l n ( 2 )}{4}} (- \frac{4 u + 19 ln ( 2 )}{ln ( 2 )} + 19)

Rewrite

- 1 = - e^{\frac{4 u + 19 l n ( 2 )}{4}} (- \frac{4 u + 19 ln ( 2 )}{ln ( 2 )} + 19)

in Lambert form:

\frac{e ^{\frac{4 u + 19 l n ( 2 )}{4}} u}{e ^{\frac{19 l n ( 2 )}{4}}} = - \frac{ln ( 2 )}{4 e ^{\frac{19 l n ( 2 )}{4}}}

Solve

\frac{e ^{\frac{4 u + 19 l n ( 2 )}{4}} u}{e ^{\frac{19 l n ( 2 )}{4}}} = - \frac{ln ( 2 )}{4 e ^{\frac{19 l n ( 2 )}{4}}} : u = W_{- 1} (- \frac{ln ( 2 )}{4 e ^{\frac{19 l n ( 2 )}{4}}}), u = W_{0} (- \frac{ln ( 2 )}{4 e ^{\frac{19 l n ( 2 )}{4}}})

u = W_{- 1} (- \frac{ln ( 2 )}{4 e ^{\frac{19 l n ( 2 )}{4}}}), u = W_{0} (- \frac{ln ( 2 )}{4 e ^{\frac{19 l n ( 2 )}{4}}})

Substitute back

u = (- x - \frac{19}{4}) ln (2),

solve for

x

Solve

(- x - \frac{19}{4}) ln (2) = W_{- 1} (- \frac{ln ( 2 )}{4 e ^{\frac{19 l n ( 2 )}{4}}}) : x = - \frac{4 W _{- 1} ( - \frac{l n ( 2 )}{4 e ^{\frac{19 l n ( 2 )}{4}}} ) + 19 ln ( 2 )}{4 ln ( 2 )}

Solve

(- x - \frac{19}{4}) ln (2) = W_{0} (- \frac{ln ( 2 )}{4 e ^{\frac{19 l n ( 2 )}{4}}}) : x = - \frac{4 W _{0} ( - \frac{l n ( 2 )}{4 e ^{\frac{19 l n ( 2 )}{4}}} ) + 19 ln ( 2 )}{4 ln ( 2 )}

x = - \frac{4 W _{- 1} ( - \frac{l n ( 2 )}{4 e ^{\frac{19 l n ( 2 )}{4}}} ) + 19 ln ( 2 )}{4 ln ( 2 )}, x = - \frac{4 W _{0} ( - \frac{l n ( 2 )}{4 e ^{\frac{19 l n ( 2 )}{4}}} ) + 19 ln ( 2 )}{4 ln ( 2 )}

e^{- 10 a - 4} = 45

14 \cdot 1 0^{0.5 w} = 100

79 = 3 5^{x}

e^{(x^{2} + 2 x + 1)} = 1

5^{3 x} = 12 5^{2 x - 3}