Solve the Inequality for x 9^(2x+1)>27

$9^{2 x + 1} > 27$

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

$3^{2 (2 x + 1)} > 3^{3}$

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

$2 (2 x + 1) > 3$

Solve for $x$ .

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

Divide each term in $2 (2 x + 1) > 3$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $2 x + 1$ by $1$ .

$2 x + 1 > \frac{3}{2}$

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

Subtract $1$ from both sides of the inequality.

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 1$ by $2$ .

$2 x > \frac{3 - 2}{2}$

Subtract $2$ from $3$ .

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

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

Divide each term in $2 x > \frac{1}{2}$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$x > \frac{1}{2} \cdot \frac{1}{2}$

Multiply $\frac{1}{2} \cdot \frac{1}{2}$ .

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

$x > \frac{1}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x > \frac{1}{4}$

The result can be shown in multiple forms.

Inequality Form:

$x > \frac{1}{4}$

Interval Notation:

$(\frac{1}{4}, \infty)$

Solve the Inequality for x 9^(2x+1)>27

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