Solve the Inequality for x 3^(-x)>6^(-x)

$3^{- x} > 6^{- x}$

Take the log of both sides of the inequality.

$\ln (3^{- x}) > \ln (6^{- x})$

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

$- x \ln (3) > \ln (6^{- x})$

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

$- x \ln (3) > - x \ln (6)$

Solve the inequality for $x$ .

Add $x \ln (6)$ to both sides of the inequality.

$- x \ln (3) + x \ln (6) > 0$

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

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

$x (- 1 \ln (3)) + x \ln (6) > 0$

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

$x (- 1 \ln (3)) + x (\ln (6)) > 0$

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

$x (- 1 \ln (3) + \ln (6)) > 0$

Rewrite $- 1 \ln (3)$ as $- \ln (3)$ .

$x (- \ln (3) + \ln (6)) > 0$

Divide each term in $x (- \ln (3) + \ln (6)) > 0$ by $- \ln (3) + \ln (6)$ and simplify.

Divide each term in $x (- \ln (3) + \ln (6)) > 0$ by $- \ln (3) + \ln (6)$ .

$\frac{x (- \ln (3) + \ln (6))}{- \ln (3) + \ln (6)} > \frac{0}{- \ln (3) + \ln (6)}$

Simplify the left side.

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

Cancel the common factor.

$\frac{x (- \ln (3) + \ln (6))}{- \ln (3) + \ln (6)} > \frac{0}{- \ln (3) + \ln (6)}$

Divide $x$ by $1$ .

$x > \frac{0}{- \ln (3) + \ln (6)}$

Simplify the right side.

Divide $0$ by $- \ln (3) + \ln (6)$ .

$x > 0$

The result can be shown in multiple forms.

Inequality Form:

$x > 0$

Interval Notation:

$(0, \infty)$