What is the answer to 1=-3/(3(0)-e^{3x)+c_{1}} ?

The answer to 1=-3/(3(0)-e^{3x)+c_{1}} is x=(ln(3+c_{1}))/3 {c_{1}>-3}

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1=-3/(3(0)-e^{3x)+c_{1}}

1 = - \frac{3}{3 ( 0 ) - e ^{3 x} + c _{1}}

x = \frac{ln ( 3 + c _{1} )}{3} {c_{1} > - 3}

Solution steps

1 = - \frac{3}{3 ( 0 ) - e ^{3 x} + c _{1}}

Multiply both sides by

c_{1} - e^{3 x}

1 \cdot (c_{1} - e^{3 x}) = - \frac{3}{3 \cdot 0 - e ^{3 x} + c _{1}} (c_{1} - e^{3 x})

Simplify

c_{1} - e^{3 x} = - 3

Move

c_{1}

to the right side

- e^{3 x} = - 3 - c_{1}

Divide both sides by

- 1

e^{3 x} = 3 + c_{1}

Apply exponent rules

3 x = ln (3 + c_{1})

Solve

3 x = ln (3 + c_{1}) : x = \frac{ln ( 3 + c _{1} )}{3}

x = \frac{ln ( 3 + c _{1} )}{3}

Verify Solutions:

x = \frac{ln ( 3 + c _{1} )}{3} {c_{1} > - 3}

The solution is

x = \frac{ln ( 3 + c _{1} )}{3} {c_{1} > - 3}

0.12 5^{x} = 32

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

3^{x + 1} - 9^{x} = 0

e^{2 x} - 9 e^{- 2 x} - 10 = 0

4^{1 - 4 x} = 3^{x}

What is the answer to 1=-3/(3(0)-e^{3x)+c_{1}} ?
The answer to 1=-3/(3(0)-e^{3x)+c_{1}} is x=(ln(3+c_{1}))/3 {c_{1}>-3}