Evaluate 1/81<9^(2x-4)

$\frac{1}{81} < 9^{2 x - 4}$

Raise $81$ to the power of $1$ .

$\frac{1}{81^{1}} < 9^{2 x - 4}$

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

$81^{- 1} < 9^{2 x - 4}$

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

$9^{- 2} < 9^{2 x - 4}$

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

$- 2 < 2 x - 4$

Solve for $x$ .

Rewrite so $x$ is on the left side of the inequality.

$2 x - 4 > - 2$

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

Add $4$ to both sides of the inequality.

$2 x > - 2 + 4$

Add $- 2$ and $4$ .

$2 x > 2$

Divide each term in $2 x > 2$ by $2$ and simplify.

Divide each term in $2 x > 2$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $2$ by $2$ .

$x > 1$

The result can be shown in multiple forms.

Inequality Form:

$x > 1$

Interval Notation:

$(1, \infty)$