What is the answer to 330e^{-0.005t}=120e^{0.035t} ?

The answer to 330e^{-0.005t}=120e^{0.035t} is t=-(ln(4/11))/(0.04)

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330e^{-0.005t}=120e^{0.035t}

330 e^{- 0.005 t} = 120 e^{0.035 t}

t = - \frac{ln ( \frac{4}{11} )}{0.04}

Decimal Notation

t = 25.29002 \dots

Solution steps

330 e^{- 0.005 t} = 120 e^{0.035 t}

Apply exponent rules

ln (330) - 0.005 t = ln (120) + 0.035 t

Solve

ln (330) - 0.005 t = ln (120) + 0.035 t : t = - \frac{ln ( \frac{4}{11} )}{0.04}

t = - \frac{ln ( \frac{4}{11} )}{0.04}

3^{x} + 1 - 2 = 25

0. 2^{x + 1} = 0.6

135 = 20 + 1130 e^{30 k}

2^{x} + 2^{4 - x} = 17

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

What is the answer to 330e^{-0.005t}=120e^{0.035t} ?
The answer to 330e^{-0.005t}=120e^{0.035t} is t=-(ln(4/11))/(0.04)