What is the answer to e^{-2t}+6=5e^{-t} ?

The answer to e^{-2t}+6=5e^{-t} is t=-ln(2),t=-ln(3)

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e^{-2t}+6=5e^{-t}

e^{- 2 t} + 6 = 5 e^{- t}

t = - ln (2), t = - ln (3)

Solution steps

e^{- 2 t} + 6 = 5 e^{- t}

Apply exponent rules

(e^{t})^{- 2} + 6 = 5 (e^{t})^{- 1}

Rewrite the equation with

e^{t} = u

(u)^{- 2} + 6 = 5 (u)^{- 1}

Solve

u^{- 2} + 6 = 5 u^{- 1} : u = \frac{1}{2}, u = \frac{1}{3}

u = \frac{1}{2}, u = \frac{1}{3}

Substitute back

u = e^{t},

solve for

t

Solve

e^{t} = \frac{1}{2} : t = - ln (2)

Solve

e^{t} = \frac{1}{3} : t = - ln (3)

t = - ln (2), t = - ln (3)

5^{x} = 40

ln^{2} (3) \cdot 3^{x} = 0

5^{x} = 49

x, 3 = 2^{x}

5^{x} = 38

What is the answer to e^{-2t}+6=5e^{-t} ?
The answer to e^{-2t}+6=5e^{-t} is t=-ln(2),t=-ln(3)