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Popular >

-9e^{-t}+17e^{2t}=0

Solution

- 9 e^{- t} + 17 e^{2 t} = 0

Solution

t = \frac{2 ln ( 3 ) - ln ( 17 )}{3}

+1

Decimal Notation

t = - 0.21199 \dots

Solution steps

- 9 e^{- t} + 17 e^{2 t} = 0

Add

9 e^{- t}

to both sides

- 9 e^{- t} + 17 e^{2 t} + 9 e^{- t} = 0 + 9 e^{- t}

Simplify

17 e^{2 t} = 9 e^{- t}

Apply exponent rules

ln (17) + 2 t = ln (9) - t

Solve

ln (17) + 2 t = ln (9) - t : t = \frac{2 ln ( 3 ) - ln ( 17 )}{3}

t = \frac{2 ln ( 3 ) - ln ( 17 )}{3}

Popular Examples

3^{0.5x-1}=(sqrt(3))/(81)

3^{0.5 x - 1} = \frac{3}{81}

3^{2 + m} = 45

5^{x} = - 4

solvefor n,2(x)^n=(2x)^n solve for

n, 2 (x)^{n} = (2 x)^{n}

x e^{3 x} - 1 = 3

Frequently Asked Questions (FAQ)

What is the answer to -9e^{-t}+17e^{2t}=0 ?
The answer to -9e^{-t}+17e^{2t}=0 is t=(2ln(3)-ln(17))/3