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Popular >

A/(1+Be^{t/2)}=C

Solution

\frac{A}{1 + B e ^{\frac{t}{2}}} = C

Solution

t = 2 ln (\frac{A - C}{BC})

Solution steps

\frac{A}{1 + B e ^{\frac{t}{2}}} = C

Multiply both sides by

1 + B e^{\frac{t}{2}}

\frac{A}{1 + B e ^{\frac{t}{2}}} (1 + B e^{\frac{t}{2}}) = C (1 + B e^{\frac{t}{2}})

Simplify

A = C (1 + B e^{\frac{t}{2}})

Switch sides

C (1 + B e^{\frac{t}{2}}) = A

Divide both sides by

C

1 + B e^{\frac{t}{2}} = \frac{A}{C}

Move

1

to the right side

B e^{\frac{t}{2}} = \frac{A}{C} - 1

Divide both sides by

B

e^{\frac{t}{2}} = \frac{A - C}{BC}

Apply exponent rules

\frac{t}{2} = ln (\frac{A - C}{BC})

Solve

\frac{t}{2} = ln (\frac{A - C}{BC}) : t = 2 ln (\frac{A - C}{BC})

t = 2 ln (\frac{A - C}{BC})

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Frequently Asked Questions (FAQ)

What is the answer to A/(1+Be^{t/2)}=C ?
The answer to A/(1+Be^{t/2)}=C is t=2ln((A-C)/(BC))