Solve the Inequality for x 3^(2x)>=6^(x+1)

$3^{2 x} \geq 6^{x + 1}$

Take the log of both sides of the inequality.

$\ln (3^{2 x}) \geq \ln (6^{x + 1})$

Expand $\ln (3^{2 x})$ by moving $2 x$ outside the logarithm.

$2 x \ln (3) \geq \ln (6^{x + 1})$

Expand $\ln (6^{x + 1})$ by moving $x + 1$ outside the logarithm.

$2 x \ln (3) \geq (x + 1) \ln (6)$

Solve the inequality for $x$ .

Simplify $(x + 1) \ln (6)$ .

Apply the distributive property.

$2 x \ln (3) \geq x \ln (6) + 1 \ln (6)$

Multiply $\ln (6)$ by $1$ .

$2 x \ln (3) \geq x \ln (6) + \ln (6)$

Subtract $x \ln (6)$ from both sides of the inequality.

$2 x \ln (3) - x \ln (6) \geq \ln (6)$

Factor $x$ out of $2 x \ln (3) - x \ln (6)$ .

Factor $x$ out of $2 x \ln (3)$ .

$x (2 \ln (3)) - x \ln (6) \geq \ln (6)$

Factor $x$ out of $- x \ln (6)$ .

$x (2 \ln (3)) + x (- 1 \ln (6)) \geq \ln (6)$

Factor $x$ out of $x (2 \ln (3)) + x (- 1 \ln (6))$ .

$x (2 \ln (3) - 1 \ln (6)) \geq \ln (6)$

Rewrite $- 1 \ln (6)$ as $- \ln (6)$ .

$x (2 \ln (3) - \ln (6)) \geq \ln (6)$

Divide each term in $x (2 \ln (3) - \ln (6)) \geq \ln (6)$ by $2 \ln (3) - \ln (6)$ and simplify.

Divide each term in $x (2 \ln (3) - \ln (6)) \geq \ln (6)$ by $2 \ln (3) - \ln (6)$ .

$\frac{x (2 \ln (3) - \ln (6))}{2 \ln (3) - \ln (6)} \geq \frac{\ln (6)}{2 \ln (3) - \ln (6)}$

Simplify the left side.

Cancel the common factor of $2 \ln (3) - \ln (6)$ .

Cancel the common factor.

$\frac{x (2 \ln (3) - \ln (6))}{2 \ln (3) - \ln (6)} \geq \frac{\ln (6)}{2 \ln (3) - \ln (6)}$

Divide $x$ by $1$ .

$x \geq \frac{\ln (6)}{2 \ln (3) - \ln (6)}$

The result can be shown in multiple forms.

Inequality Form:

$x \geq \frac{\ln (6)}{2 \ln (3) - \ln (6)}$

Interval Notation:

$[\frac{\ln (6)}{2 \ln (3) - \ln (6)}, \infty)$