Solve for t H=a+be^(-kt)

$H = a + b e^{- k t}$

Rewrite the equation as $a + b e^{- k t} = H$ .

$a + b e^{- k t} = H$

Subtract $a$ from both sides of the equation.

$b e^{- k t} = H - a$

Divide each term by $b$ and simplify.

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Divide each term in $b e^{- k t} = H - a$ by $b$ .

$\frac{b e^{- k t}}{b} = \frac{H}{b} + \frac{- a}{b}$

Cancel the common factor of $b$ .

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Cancel the common factor.

$\frac{b e^{- k t}}{b} = \frac{H}{b} + \frac{- a}{b}$

Divide $e^{- k t}$ by $1$ .

$e^{- k t} = \frac{H}{b} + \frac{- a}{b}$

Move the negative in front of the fraction.

$e^{- k t} = \frac{H}{b} - \frac{a}{b}$

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{- k t}) = \ln (\frac{H}{b} - \frac{a}{b})$

Expand the left side.

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Expand $\ln (e^{- k t})$ by moving $- k t$ outside the logarithm.

$- k t \ln (e) = \ln (\frac{H}{b} - \frac{a}{b})$

The natural logarithm of $e$ is $1$ .

$- k t \cdot 1 = \ln (\frac{H}{b} - \frac{a}{b})$

Multiply $- 1$ by $1$ .

$- k t = \ln (\frac{H}{b} - \frac{a}{b})$

Divide each term by $- 1 k$ and simplify.

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Divide each term in $- k t = \ln (\frac{H}{b} - \frac{a}{b})$ by $- 1 k$ .

$\frac{- k t}{- 1 k} = \frac{\ln (\frac{H}{b} - \frac{a}{b})}{- 1 k}$

Simplify $\frac{- k t}{- 1 k}$ .

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Dividing two negative values results in a positive value.

$\frac{k t}{k} = \frac{\ln (\frac{H}{b} - \frac{a}{b})}{- 1 k}$

Cancel the common factor of $k$ .

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Cancel the common factor.

$\frac{k t}{k} = \frac{\ln (\frac{H}{b} - \frac{a}{b})}{- 1 k}$

Divide $t$ by $1$ .

$t = \frac{\ln (\frac{H}{b} - \frac{a}{b})}{- 1 k}$

Move the negative in front of the fraction.

$t = - \frac{\ln (\frac{H}{b} - \frac{a}{b})}{k}$

Solve for t H=a+be^(-kt)

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