Solve for n 3^(2n)>=6^(n+1)

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

Step 1

Take the log of both sides of the inequality.

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

Step 2

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

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

Step 3

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

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

Step 4

Solve the inequality for $n$ .

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

Apply the distributive property.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Simplify the left side.

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

Cancel the common factor.

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

Divide $n$ by $1$ .

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

Step 5

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

Step 6