Solve for t natural log of A = natural log of C-kt

$\ln (A) = \ln (C) - k t$

Rewrite the equation as $\ln (C) - k t = \ln (A)$ .

$\ln (C) - k t = \ln (A)$

Move all the terms containing a logarithm to the left side of the equation.

$\ln (C) - \ln (A) = k t$

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{C}{A}) = k t$

Since $t$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$k t = \ln (\frac{C}{A})$

Divide each term in $k t = \ln (\frac{C}{A})$ by $k$ and simplify.

Divide each term in $k t = \ln (\frac{C}{A})$ by $k$ .

$\frac{k t}{k} = \frac{\ln (\frac{C}{A})}{k}$

Simplify the left side.

Cancel the common factor of $k$ .

Cancel the common factor.

$\frac{k t}{k} = \frac{\ln (\frac{C}{A})}{k}$

Divide $t$ by $1$ .

$t = \frac{\ln (\frac{C}{A})}{k}$