Calculus Examples

$e^{2 - 3 x} > 4$

Step 1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{2 - 3 x}) > \ln (4)$

Step 2

Expand the left side.

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Step 2.1

Expand $\ln (e^{2 - 3 x})$ by moving $2 - 3 x$ outside the logarithm.

$(2 - 3 x) \ln (e) > \ln (4)$

Step 2.2

The natural logarithm of $e$ is $1$ .

$(2 - 3 x) \cdot 1 > \ln (4)$

Step 2.3

Multiply $2 - 3 x$ by $1$ .

$2 - 3 x > \ln (4)$

Step 3

Subtract $2$ from both sides of the inequality.

$- 3 x > \ln (4) - 2$

Step 4

Divide each term in $- 3 x > \ln (4) - 2$ by $- 3$ and simplify.

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Step 4.1

Divide each term in $- 3 x > \ln (4) - 2$ by $- 3$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 3 x}{- 3} < \frac{\ln (4)}{- 3} + \frac{- 2}{- 3}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $- 3$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{- 3 x}{- 3} < \frac{\ln (4)}{- 3} + \frac{- 2}{- 3}$

Step 4.2.1.2

Divide $x$ by $1$ .

$x < \frac{\ln (4)}{- 3} + \frac{- 2}{- 3}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Simplify each term.

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Step 4.3.1.1

Move the negative in front of the fraction.

$x < - \frac{\ln (4)}{3} + \frac{- 2}{- 3}$

Step 4.3.1.2

Dividing two negative values results in a positive value.

$x < - \frac{\ln (4)}{3} + \frac{2}{3}$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x < - \frac{\ln (4)}{3} + \frac{2}{3}$

Interval Notation:

$(- \infty, - \frac{\ln (4)}{3} + \frac{2}{3})$

Step 6