What is the answer to 2=ce^{-t}+1/(2e^t) ?

The answer to 2=ce^{-t}+1/(2e^t) is t=-ln(4/(2c+1)) {c -1/2 }

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2=ce^{-t}+1/(2e^t)

2 = c e^{- t} + \frac{1}{2 e ^{t}}

t = - ln (\frac{4}{2 c + 1}) {c < - \frac{1}{2} or c > - \frac{1}{2}}

Solution steps

2 = c e^{- t} + \frac{1}{2 e ^{t}}

Apply exponent rules

2 = c e^{- t} + \frac{e ^{- t}}{2}

Multiply both sides by

2

2 \cdot 2 = c e^{- t} \cdot 2 + \frac{e ^{- t}}{2} \cdot 2

Simplify

4 = 2 c e^{- t} + e^{- t}

Switch sides

2 c e^{- t} + e^{- t} = 4

Factor

2 c e^{- t} + e^{- t} : e^{- t} (2 c + 1)

e^{- t} (2 c + 1) = 4

Divide both sides by

2 c + 1

e^{- t} = \frac{4}{2 c + 1}

Apply exponent rules

- t = ln (\frac{4}{2 c + 1})

Solve

- t = ln (\frac{4}{2 c + 1}) : t = - ln (\frac{4}{2 c + 1})

t = - ln (\frac{4}{2 c + 1})

Verify Solutions:

t = - ln (\frac{4}{2 c + 1}) {c < - \frac{1}{2} or c > - \frac{1}{2}}

The solution is

t = - ln (\frac{4}{2 c + 1}) {c < - \frac{1}{2} or c > - \frac{1}{2}}

8^{5 x} = 1 6^{3 x + 4}

2^{x - 1} = 22

e^{4 x} - 5 e^{2 x} + 4 = 0

x, e^{x^{2} + y^{2}} = 3

\frac{e ^{x} - 1}{x} = 1

What is the answer to 2=ce^{-t}+1/(2e^t) ?
The answer to 2=ce^{-t}+1/(2e^t) is t=-ln(4/(2c+1)) {c<-1/2 \lor c>-1/2 }