What is the answer to 2=2^t-2t ?

The answer to 2=2^t-2t is t=3,t=-(\W_{0}(-(ln(2))/4)+ln(2))/(ln(2))

Solve for t: 2=2^t-2t

The solution is t=3,t=-(\W_{0}(-(ln(2))/4)+ln(2))/(ln(2))

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2 = 2^{t} - 2 t

t = 3, t = - \frac{W _{0} ( - \frac{l n ( 2 )}{4} ) + ln ( 2 )}{ln ( 2 )}

Decimal Notation

t = 3, t = - 0.69009 \dots

Solution steps

2 = 2^{t} - 2 t

Prepare

2 = 2^{t} - 2 t

for Lambert form:

1 = (2 + 2 t) e^{- l n (2) t}

Rewrite the equation with

(- t - 1) ln (2) = u

and

t = - \frac{u + ln ( 2 )}{ln ( 2 )}

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

Simplify

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

Rewrite

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

in Lambert form:

e^{u} u = - \frac{ln ( 2 )}{4}

Solve

e^{u} u = - \frac{ln ( 2 )}{4} : u = - 4 ln (2), u = W_{0} (- \frac{ln ( 2 )}{4})

u = - 4 ln (2), u = W_{0} (- \frac{ln ( 2 )}{4})

Substitute back

u = (- t - 1) ln (2),

solve for

t

Solve

(- t - 1) ln (2) = - 4 ln (2) : t = 3

Solve

(- t - 1) ln (2) = W_{0} (- \frac{ln ( 2 )}{4}) : t = - \frac{W _{0} ( - \frac{l n ( 2 )}{4} ) + ln ( 2 )}{ln ( 2 )}

t = 3, t = - \frac{W _{0} ( - \frac{l n ( 2 )}{4} ) + ln ( 2 )}{ln ( 2 )}

\frac{1}{3 ^{x} + 3 ^{x} + 3 ^{x}} = 81

6 x e^{x} + 5 e^{x + 2} = 0

3^{2 x - 3} = 5^{4 x + 5}

3^{x} = 5^{x - 1}

27 = 3^{5 x} 9^{x^{2}}

What is the answer to 2=2^t-2t ?
The answer to 2=2^t-2t is t=3,t=-(\W_{0}(-(ln(2))/4)+ln(2))/(ln(2))
Solve for t: 2=2^t-2t
The solution is t=3,t=-(\W_{0}(-(ln(2))/4)+ln(2))/(ln(2))