Precalculus Examples

$r = p - k \ln (t)$

Step 1

Rewrite the equation as $p - k \ln (t) = r$ .

$p - k \ln (t) = r$

Step 2

Subtract $p$ from both sides of the equation.

$- k \ln (t) = r - p$

Step 3

Divide each term in $- k \ln (t) = r - p$ by $- k$ and simplify.

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Step 3.1

Divide each term in $- k \ln (t) = r - p$ by $- k$ .

$\frac{- k \ln (t)}{- k} = \frac{r}{- k} + \frac{- p}{- k}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Dividing two negative values results in a positive value.

$\frac{k \ln (t)}{k} = \frac{r}{- k} + \frac{- p}{- k}$

Step 3.2.2

Cancel the common factor of $k$ .

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Step 3.2.2.1

Cancel the common factor.

$\frac{k \ln (t)}{k} = \frac{r}{- k} + \frac{- p}{- k}$

Step 3.2.2.2

Divide $\ln (t)$ by $1$ .

$\ln (t) = \frac{r}{- k} + \frac{- p}{- k}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Simplify each term.

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Step 3.3.1.1

Move the negative in front of the fraction.

$\ln (t) = - \frac{r}{k} + \frac{- p}{- k}$

Step 3.3.1.2

Dividing two negative values results in a positive value.

$\ln (t) = - \frac{r}{k} + \frac{p}{k}$

Step 4

To solve for $t$ , rewrite the equation using properties of logarithms.

$e^{\ln (t)} = e^{- \frac{r}{k} + \frac{p}{k}}$

Step 5

Rewrite $\ln (t) = - \frac{r}{k} + \frac{p}{k}$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{- \frac{r}{k} + \frac{p}{k}} = t$

Step 6

Rewrite the equation as $t = e^{- \frac{r}{k} + \frac{p}{k}}$ .

$t = e^{- \frac{r}{k} + \frac{p}{k}}$