Solve for x 4^(2x)+1<=65

$4^{2 x} + 1 \leq 65$

Step 1

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

Subtract $1$ from both sides of the inequality.

$4^{2 x} \leq 65 - 1$

Subtract $1$ from $65$ .

$4^{2 x} \leq 64$

Step 2

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

$4^{2 x} \leq 4^{3}$

Step 3

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

$2 x \leq 3$

Step 4

Divide each term in $2 x \leq 3$ by $2$ and simplify.

Divide each term in $2 x \leq 3$ by $2$ .

$\frac{2 x}{2} \leq \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \leq \frac{3}{2}$

Divide $x$ by $1$ .

$x \leq \frac{3}{2}$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x \leq \frac{3}{2}$

Interval Notation:

$(- \infty, \frac{3}{2}]$

Step 6