Solve the Inequality for x (1/16)^(3x-4)<=64^(x-1)

${(\frac{1}{16})}^{3 x - 4} \leq 64^{x - 1}$

Apply the product rule to $\frac{1}{16}$ .

$\frac{1^{3 x - 4}}{16^{3 x - 4}} \leq 64^{x - 1}$

One to any power is one.

$\frac{1}{16^{3 x - 4}} \leq 64^{x - 1}$

Move $16^{3 x - 4}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$16^{- (3 x - 4)} \leq 64^{x - 1}$

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

$4^{2 (- (3 x - 4))} \leq 4^{3 (x - 1)}$

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

$2 (- (3 x - 4)) \leq 3 (x - 1)$

Solve for $x$ .

Simplify $2 (- (3 x - 4))$ .

Start simplifying.

$0 + 0 + 2 (- (3 x - 4)) \leq 3 (x - 1)$

Simplify by adding zeros.

$2 (- (3 x - 4)) \leq 3 (x - 1)$

Apply the distributive property.

$2 (- (3 x) - - 4) \leq 3 (x - 1)$

Multiply.

Multiply $3$ by $- 1$ .

$2 (- 3 x - - 4) \leq 3 (x - 1)$

Multiply $- 1$ by $- 4$ .

$2 (- 3 x + 4) \leq 3 (x - 1)$

Apply the distributive property.

$2 (- 3 x) + 2 \cdot 4 \leq 3 (x - 1)$

Multiply.

Multiply $- 3$ by $2$ .

$- 6 x + 2 \cdot 4 \leq 3 (x - 1)$

Multiply $2$ by $4$ .

$- 6 x + 8 \leq 3 (x - 1)$

Simplify $3 (x - 1)$ .

Apply the distributive property.

$- 6 x + 8 \leq 3 x + 3 \cdot - 1$

Multiply $3$ by $- 1$ .

$- 6 x + 8 \leq 3 x - 3$

Move all terms containing $x$ to the left side of the inequality.

Subtract $3 x$ from both sides of the inequality.

$- 6 x + 8 - 3 x \leq - 3$

Subtract $3 x$ from $- 6 x$ .

$- 9 x + 8 \leq - 3$

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

Subtract $8$ from both sides of the inequality.

$- 9 x \leq - 3 - 8$

Subtract $8$ from $- 3$ .

$- 9 x \leq - 11$

Divide each term in $- 9 x \leq - 11$ by $- 9$ and simplify.

Divide each term in $- 9 x \leq - 11$ by $- 9$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 9 x}{- 9} \geq \frac{- 11}{- 9}$

Simplify the left side.

Cancel the common factor of $- 9$ .

Cancel the common factor.

$\frac{- 9 x}{- 9} \geq \frac{- 11}{- 9}$

Divide $x$ by $1$ .

$x \geq \frac{- 11}{- 9}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x \geq \frac{11}{9}$

The result can be shown in multiple forms.

Inequality Form:

$x \geq \frac{11}{9}$

Interval Notation:

$[\frac{11}{9}, \infty)$

Solve the Inequality for x (1/16)^(3x-4)<=64^(x-1)

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