Solve the Inequality for x (1/81)^(3-x)<=9

${(\frac{1}{81})}^{3 - x} \leq 9$

Apply the product rule to $\frac{1}{81}$ .

$\frac{1^{3 - x}}{81^{3 - x}} \leq 9$

One to any power is one.

$\frac{1}{81^{3 - x}} \leq 9$

Move $81^{3 - x}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$81^{- (3 - x)} \leq 9$

Create equivalent expressions in the equation that all have equal bases.

$9^{2 (- (3 - x))} \leq 9^{1}$

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$2 (- (3 - x)) \leq 1$

Solve for $x$ .

Simplify.

Apply the distributive property.

$2 (- 1 \cdot 3 - - x) \leq 1$

Multiply $- 1$ by $3$ .

$2 (- 3 - - x) \leq 1$

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

$2 (- 3 + 1 x) \leq 1$

Multiply $x$ by $1$ .

$2 (- 3 + x) \leq 1$

Divide each term in $2 (- 3 + x) \leq 1$ by $- 2$ and simplify.

Divide each term in $2 (- 3 + x) \leq 1$ by $- 2$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{2 (- 3 + x)}{- 2} \geq \frac{1}{- 2}$

Simplify the left side.

Simplify the expression.

Move the negative one from the denominator of $\frac{- 3 + x}{- 1}$ .

$- 1 \cdot (- 3 + x) \geq \frac{1}{- 2}$

Rewrite $- 1 \cdot (- 3 + x)$ as $- (- 3 + x)$ .

$- (- 3 + x) \geq \frac{1}{- 2}$

Apply the distributive property.

$- - 3 - x \geq \frac{1}{- 2}$

Multiply $- 1$ by $- 3$ .

$3 - x \geq \frac{1}{- 2}$

Simplify the right side.

Move the negative in front of the fraction.

$3 - x \geq - \frac{1}{2}$

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

Subtract $3$ from both sides of the inequality.

$- x \geq - \frac{1}{2} - 3$

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- x \geq - \frac{1}{2} - 3 \cdot \frac{2}{2}$

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

$- x \geq - \frac{1}{2} + \frac{- 3 \cdot 2}{2}$

Combine the numerators over the common denominator.

$- x \geq \frac{- 1 - 3 \cdot 2}{2}$

Simplify the numerator.

Multiply $- 3$ by $2$ .

$- x \geq \frac{- 1 - 6}{2}$

Subtract $6$ from $- 1$ .

$- x \geq \frac{- 7}{2}$

Move the negative in front of the fraction.

$- x \geq - \frac{7}{2}$

Divide each term in $- x \geq - \frac{7}{2}$ by $- 1$ and simplify.

Divide each term in $- x \geq - \frac{7}{2}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{- \frac{7}{2}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{- \frac{7}{2}}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{- \frac{7}{2}}{- 1}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x \leq \frac{\frac{7}{2}}{1}$

Divide $\frac{7}{2}$ by $1$ .

$x \leq \frac{7}{2}$

The result can be shown in multiple forms.

Inequality Form:

$x \leq \frac{7}{2}$

Interval Notation:

$(- \infty, \frac{7}{2}]$

Solve the Inequality for x (1/81)^(3-x)<=9

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