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

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

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

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

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

Decimal Notation

t = 25.29002 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

x, 8 y = 3 e^{x}

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

e^{(2 x + 3)} = 37

8 (2^{2 x}) + 31 (2^{x}) - 4 = 0

(2^{4})^{3} = 8^{x}

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