Solve for x 16^(x+1)>=4^(4x+1)

$16^{x + 1} \geq 4^{4 x + 1}$

Step 1

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

$4^{2 (x + 1)} \geq 4^{4 x + 1}$

Step 2

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

$2 (x + 1) \geq 4 x + 1$

Step 3

Solve for $x$ .

Simplify $2 (x + 1)$ .

Rewrite.

$0 + 0 + 2 (x + 1) \geq 4 x + 1$

Simplify by adding zeros.

$2 (x + 1) \geq 4 x + 1$

Apply the distributive property.

$2 x + 2 \cdot 1 \geq 4 x + 1$

Multiply $2$ by $1$ .

$2 x + 2 \geq 4 x + 1$

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

Subtract $4 x$ from both sides of the inequality.

$2 x + 2 - 4 x \geq 1$

Subtract $4 x$ from $2 x$ .

$- 2 x + 2 \geq 1$

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

Subtract $2$ from both sides of the inequality.

$- 2 x \geq 1 - 2$

Subtract $2$ from $1$ .

$- 2 x \geq - 1$

Divide each term in $- 2 x \geq - 1$ by $- 2$ and simplify.

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

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

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Dividing two negative values results in a positive value.

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

Step 4

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

$(- \infty, \frac{1}{2}]$

Step 5

Solve for x 16^(x+1)>=4^(4x+1)

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