Precalculus Examples

$5 \ln (3 - x) = 4$

Step 1

Divide each term in $5 \ln (3 - x) = 4$ by $5$ and simplify.

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Step 1.1

Divide each term in $5 \ln (3 - x) = 4$ by $5$ .

$\frac{5 \ln (3 - x)}{5} = \frac{4}{5}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of $5$ .

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Step 1.2.1.1

Cancel the common factor.

$\frac{5 \ln (3 - x)}{5} = \frac{4}{5}$

Step 1.2.1.2

Divide $\ln (3 - x)$ by $1$ .

$\ln (3 - x) = \frac{4}{5}$

Step 2

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

$e^{\ln (3 - x)} = e^{\frac{4}{5}}$

Step 3

Rewrite $\ln (3 - x) = \frac{4}{5}$ 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 $b^{y} = x$ .

$e^{\frac{4}{5}} = 3 - x$

Step 4

Solve for $x$ .

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Step 4.1

Rewrite the equation as $3 - x = e^{\frac{4}{5}}$ .

$3 - x = e^{\frac{4}{5}}$

Step 4.2

Subtract $3$ from both sides of the equation.

$- x = e^{\frac{4}{5}} - 3$

Step 4.3

Divide each term in $- x = e^{\frac{4}{5}} - 3$ by $- 1$ and simplify.

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Step 4.3.1

Divide each term in $- x = e^{\frac{4}{5}} - 3$ by $- 1$ .

$\frac{- x}{- 1} = \frac{e^{\frac{4}{5}}}{- 1} + \frac{- 3}{- 1}$

Step 4.3.2

Simplify the left side.

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Step 4.3.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{e^{\frac{4}{5}}}{- 1} + \frac{- 3}{- 1}$

Step 4.3.2.2

Divide $x$ by $1$ .

$x = \frac{e^{\frac{4}{5}}}{- 1} + \frac{- 3}{- 1}$

Step 4.3.3

Simplify the right side.

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Step 4.3.3.1

Simplify each term.

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Step 4.3.3.1.1

Move the negative one from the denominator of $\frac{e^{\frac{4}{5}}}{- 1}$ .

$x = - 1 \cdot e^{\frac{4}{5}} + \frac{- 3}{- 1}$

Step 4.3.3.1.2

Rewrite $- 1 \cdot e^{\frac{4}{5}}$ as $- e^{\frac{4}{5}}$ .

$x = - e^{\frac{4}{5}} + \frac{- 3}{- 1}$

Step 4.3.3.1.3

Divide $- 3$ by $- 1$ .

$x = - e^{\frac{4}{5}} + 3$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - e^{\frac{4}{5}} + 3$

Decimal Form:

$x = 0.77445907 \dots$