Precalculus Examples

$\ln (2 - 5 x) > 2$

Step 1

Convert the inequality to an equality.

$\ln (2 - 5 x) = 2$

Step 2

Solve the equation.

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Step 2.1

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

$e^{\ln (2 - 5 x)} = e^{2}$

Step 2.2

Rewrite $\ln (2 - 5 x) = 2$ 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 $b^{y} = x$ .

$e^{2} = 2 - 5 x$

Step 2.3

Solve for $x$ .

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Step 2.3.1

Rewrite the equation as $2 - 5 x = e^{2}$ .

$2 - 5 x = e^{2}$

Step 2.3.2

Subtract $2$ from both sides of the equation.

$- 5 x = e^{2} - 2$

Step 2.3.3

Divide each term in $- 5 x = e^{2} - 2$ by $- 5$ and simplify.

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Step 2.3.3.1

Divide each term in $- 5 x = e^{2} - 2$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{e^{2}}{- 5} + \frac{- 2}{- 5}$

Step 2.3.3.2

Simplify the left side.

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Step 2.3.3.2.1

Cancel the common factor of $- 5$ .

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Step 2.3.3.2.1.1

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{e^{2}}{- 5} + \frac{- 2}{- 5}$

Step 2.3.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{e^{2}}{- 5} + \frac{- 2}{- 5}$

Step 2.3.3.3

Simplify the right side.

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Step 2.3.3.3.1

Simplify each term.

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Step 2.3.3.3.1.1

Move the negative in front of the fraction.

$x = - \frac{e^{2}}{5} + \frac{- 2}{- 5}$

Step 2.3.3.3.1.2

Dividing two negative values results in a positive value.

$x = - \frac{e^{2}}{5} + \frac{2}{5}$

Step 3

Find the domain of $\ln (2 - 5 x) - 2$ .

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Step 3.1

Set the argument in $\ln (2 - 5 x)$ greater than $0$ to find where the expression is defined.

$2 - 5 x > 0$

Step 3.2

Solve for $x$ .

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Step 3.2.1

Subtract $2$ from both sides of the inequality.

$- 5 x > - 2$

Step 3.2.2

Divide each term in $- 5 x > - 2$ by $- 5$ and simplify.

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Step 3.2.2.1

Divide each term in $- 5 x > - 2$ by $- 5$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 5 x}{- 5} < \frac{- 2}{- 5}$

Step 3.2.2.2

Simplify the left side.

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Step 3.2.2.2.1

Cancel the common factor of $- 5$ .

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Step 3.2.2.2.1.1

Cancel the common factor.

$\frac{- 5 x}{- 5} < \frac{- 2}{- 5}$

Step 3.2.2.2.1.2

Divide $x$ by $1$ .

$x < \frac{- 2}{- 5}$

Step 3.2.2.3

Simplify the right side.

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Step 3.2.2.3.1

Dividing two negative values results in a positive value.

$x < \frac{2}{5}$

Step 3.3

The domain is all values of $x$ that make the expression defined.

$(- \infty, \frac{2}{5})$

Step 4

The solution consists of all of the true intervals.

$x < - \frac{e^{2}}{5} + \frac{2}{5}$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x < - \frac{e^{2}}{5} + \frac{2}{5}$

Interval Notation:

$(- \infty, - \frac{e^{2}}{5} + \frac{2}{5})$

Step 6