Calculus Examples

$1 - \ln (1 - x) > 0$

Step 1

Convert the inequality to an equality.

$1 - \ln (1 - x) = 0$

Step 2

Solve the equation.

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Step 2.1

Subtract $1$ from both sides of the equation.

$- \ln (1 - x) = - 1$

Step 2.2

Divide each term in $- \ln (1 - x) = - 1$ by $- 1$ and simplify.

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Step 2.2.1

Divide each term in $- \ln (1 - x) = - 1$ by $- 1$ .

$\frac{- \ln (1 - x)}{- 1} = \frac{- 1}{- 1}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Dividing two negative values results in a positive value.

$\frac{\ln (1 - x)}{1} = \frac{- 1}{- 1}$

Step 2.2.2.2

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

$\ln (1 - x) = \frac{- 1}{- 1}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $- 1$ by $- 1$ .

$\ln (1 - x) = 1$

Step 2.3

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

$e^{\ln (1 - x)} = e^{1}$

Step 2.4

Rewrite $\ln (1 - x) = 1$ 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 $b^{y} = x$ .

$e^{1} = 1 - x$

Step 2.5

Solve for $x$ .

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Step 2.5.1

Rewrite the equation as $1 - x = e^{1}$ .

$1 - x = e$

Step 2.5.2

Subtract $1$ from both sides of the equation.

$- x = e - 1$

Step 2.5.3

Divide each term in $- x = e - 1$ by $- 1$ and simplify.

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Step 2.5.3.1

Divide each term in $- x = e - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{e}{- 1} + \frac{- 1}{- 1}$

Step 2.5.3.2

Simplify the left side.

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Step 2.5.3.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{e}{- 1} + \frac{- 1}{- 1}$

Step 2.5.3.2.2

Divide $x$ by $1$ .

$x = \frac{e}{- 1} + \frac{- 1}{- 1}$

Step 2.5.3.3

Simplify the right side.

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Step 2.5.3.3.1

Simplify each term.

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Step 2.5.3.3.1.1

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

$x = - 1 \cdot e + \frac{- 1}{- 1}$

Step 2.5.3.3.1.2

Rewrite $- 1 \cdot e$ as $- e$ .

$x = - e + \frac{- 1}{- 1}$

Step 2.5.3.3.1.3

Divide $- 1$ by $- 1$ .

$x = - e + 1$

Step 3

Find the domain of $1 - \ln (1 - x)$ .

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Step 3.1

Set the argument in $\ln (1 - x)$ greater than $0$ to find where the expression is defined.

$1 - x > 0$

Step 3.2

Solve for $x$ .

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Step 3.2.1

Subtract $1$ from both sides of the inequality.

$- x > - 1$

Step 3.2.2

Divide each term in $- x > - 1$ by $- 1$ and simplify.

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Step 3.2.2.1

Divide each term in $- x > - 1$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} < \frac{- 1}{- 1}$

Step 3.2.2.2

Simplify the left side.

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Step 3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} < \frac{- 1}{- 1}$

Step 3.2.2.2.2

Divide $x$ by $1$ .

$x < \frac{- 1}{- 1}$

Step 3.2.2.3

Simplify the right side.

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Step 3.2.2.3.1

Divide $- 1$ by $- 1$ .

$x < 1$

Step 3.3

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

$(- \infty, 1)$

Step 4

Use each root to create test intervals.

$x < - e + 1$

$- e + 1 < x < 1$

$x > 1$

Step 5

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 5.1

Test a value on the interval $x < - e + 1$ to see if it makes the inequality true.

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Step 5.1.1

Choose a value on the interval $x < - e + 1$ and see if this value makes the original inequality true.

$x = - 4$

Step 5.1.2

Replace $x$ with $- 4$ in the original inequality.

$1 - \ln (1 - (- 4)) > 0$

Step 5.1.3

The left side $- 0.60943791$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 5.2

Test a value on the interval $- e + 1 < x < 1$ to see if it makes the inequality true.

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Step 5.2.1

Choose a value on the interval $- e + 1 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 5.2.2

Replace $x$ with $0$ in the original inequality.

$1 - \ln (1 - (0)) > 0$

Step 5.2.3

The left side $1$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 5.3

Test a value on the interval $x > 1$ to see if it makes the inequality true.

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Step 5.3.1

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Step 5.3.2

Replace $x$ with $4$ in the original inequality.

$1 - \ln (1 - (4)) > 0$

Step 5.3.3

Determine if the inequality is true.

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Step 5.3.3.1

The equation cannot be solved because it is undefined.

$Undefined$

Step 5.3.3.2

The left side has no solution, which means that the given statement is false.

False

Step 5.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < - e + 1$ False

$- e + 1 < x < 1$ True

$x > 1$ False

$x < - e + 1$ False

$- e + 1 < x < 1$ True

$x > 1$ False

Step 6

The solution consists of all of the true intervals.

$- e + 1 < x < 1$

Step 7

The result can be shown in multiple forms.

Inequality Form:

$- e + 1 < x < 1$

Interval Notation:

$(- e + 1, 1)$

Step 8