Solve the Inequality for x 2/3(1-e^(-x))<=-3

$\frac{2}{3} (1 - e^{- x}) \leq - 3$

Simplify.

Apply the distributive property.

$\frac{2}{3} \cdot 1 + \frac{2}{3} (- e^{- x}) \leq - 3$

Multiply $\frac{2}{3}$ by $1$ .

$\frac{2}{3} + \frac{2}{3} (- e^{- x}) \leq - 3$

Combine $\frac{2}{3}$ and $e^{- x}$ .

$\frac{2}{3} - \frac{2 e^{- x}}{3} \leq - 3$

Multiply each term in $\frac{2}{3} - \frac{2 e^{- x}}{3} \leq - 3$ by $\frac{3}{2}$ to eliminate the fractions.

Multiply each term in $\frac{2}{3} - \frac{2 e^{- x}}{3} \leq - 3$ by $\frac{3}{2}$ .

$\frac{2}{3} \cdot \frac{3}{2} - \frac{2 e^{- x}}{3} \cdot \frac{3}{2} \leq - 3 (\frac{3}{2})$

Simplify the left side.

Simplify each term.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2}{3} \cdot \frac{3}{2} - \frac{2 e^{- x}}{3} \cdot \frac{3}{2} \leq - 3 (\frac{3}{2})$

Rewrite the expression.

$\frac{1}{3} \cdot 3 - \frac{2 e^{- x}}{3} \cdot \frac{3}{2} \leq - 3 (\frac{3}{2})$

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{1}{3} \cdot 3 - \frac{2 e^{- x}}{3} \cdot \frac{3}{2} \leq - 3 (\frac{3}{2})$

Rewrite the expression.

$1 - \frac{2 e^{- x}}{3} \cdot \frac{3}{2} \leq - 3 (\frac{3}{2})$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{2 e^{- x}}{3}$ into the numerator.

$1 + \frac{- 2 e^{- x}}{3} \cdot \frac{3}{2} \leq - 3 (\frac{3}{2})$

Factor $2$ out of $- 2 e^{- x}$ .

$1 + \frac{2 (- e^{- x})}{3} \cdot \frac{3}{2} \leq - 3 (\frac{3}{2})$

Cancel the common factor.

$1 + \frac{2 (- e^{- x})}{3} \cdot \frac{3}{2} \leq - 3 (\frac{3}{2})$

Rewrite the expression.

$1 + \frac{- e^{- x}}{3} \cdot 3 \leq - 3 (\frac{3}{2})$

Cancel the common factor of $3$ .

Cancel the common factor.

$1 + \frac{- e^{- x}}{3} \cdot 3 \leq - 3 (\frac{3}{2})$

Rewrite the expression.

$1 - e^{- x} \leq - 3 (\frac{3}{2})$

Simplify the right side.

Multiply $- 3 (\frac{3}{2})$ .

Combine $- 3$ and $\frac{3}{2}$ .

$1 - e^{- x} \leq \frac{- 3 \cdot 3}{2}$

Multiply $- 3$ by $3$ .

$1 - e^{- x} \leq \frac{- 9}{2}$

Move the negative in front of the fraction.

$1 - e^{- x} \leq - \frac{9}{2}$

Move all terms not containing $x$ to the right side of the inequality.

Subtract $1$ from both sides of the inequality.

$- e^{- x} \leq - \frac{9}{2} - 1$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- e^{- x} \leq - \frac{9}{2} - 1 \cdot \frac{2}{2}$

Combine $- 1$ and $\frac{2}{2}$ .

$- e^{- x} \leq - \frac{9}{2} + \frac{- 1 \cdot 2}{2}$

Combine the numerators over the common denominator.

$- e^{- x} \leq \frac{- 9 - 1 \cdot 2}{2}$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$- e^{- x} \leq \frac{- 9 - 2}{2}$

Subtract $2$ from $- 9$ .

$- e^{- x} \leq \frac{- 11}{2}$

Move the negative in front of the fraction.

$- e^{- x} \leq - \frac{11}{2}$

Divide each term in $- e^{- x} \leq - \frac{11}{2}$ by $- 1$ and simplify.

Divide each term in $- e^{- x} \leq - \frac{11}{2}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- e^{- x}}{- 1} \geq \frac{- \frac{11}{2}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{e^{- x}}{1} \geq \frac{- \frac{11}{2}}{- 1}$

Divide $e^{- x}$ by $1$ .

$e^{- x} \geq \frac{- \frac{11}{2}}{- 1}$

Simplify the right side.

Dividing two negative values results in a positive value.

$e^{- x} \geq \frac{\frac{11}{2}}{1}$

Divide $\frac{11}{2}$ by $1$ .

$e^{- x} \geq \frac{11}{2}$

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

$\ln (e^{- x}) \geq \ln (\frac{11}{2})$

Expand the left side.

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

$- x \ln (e) \geq \ln (\frac{11}{2})$

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

$- x \cdot 1 \geq \ln (\frac{11}{2})$

Multiply $- 1$ by $1$ .

$- x \geq \ln (\frac{11}{2})$

Divide each term in $- x \geq \ln (\frac{11}{2})$ by $- 1$ and simplify.

Divide each term in $- x \geq \ln (\frac{11}{2})$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{\ln (\frac{11}{2})}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{\ln (\frac{11}{2})}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{\ln (\frac{11}{2})}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\ln (\frac{11}{2})}{- 1}$ .

$x \leq - 1 \cdot \ln (\frac{11}{2})$

Rewrite $- 1 \cdot \ln (\frac{11}{2})$ as $- \ln (\frac{11}{2})$ .

$x \leq - \ln (\frac{11}{2})$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - \ln (\frac{11}{2})$

Interval Notation:

$(- \infty, - \ln (\frac{11}{2})]$

Solve the Inequality for x 2/3(1-e^(-x))<=-3

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