To solve the equation $3^{3x^{-2}} = 31$, we need to isolate $x$.

1. Start by taking the natural logarithm of both sides: $\ln(3^{3x^{-2}}) = \ln(31)$

2. Use the power rule of logarithms, $\ln(a^b) = b \ln(a)$, to bring the exponent down: $3x^{-2} \cdot \ln(3) = \ln(31)$

3. Solve for $x^{-2}$ by dividing both sides by $3 \ln(3)$: $x^{-2} = \frac{\ln(31)}{3 \ln(3)}$

4. Now take the reciprocal to solve for $x^2$: $x^2 = \frac{3 \ln(3)}{\ln(31)}$

5. Finally, take the square root of both sides to solve for $x$: $x = \pm \sqrt{\frac{3 \ln(3)}{\ln(31)}}$

Numerical Solution

Calculating this to four decimal places:

$x \approx \pm 0.9246$

Thus, the solution for $x$ is approximately: $x \approx \pm 0.9246$

Would you like a step-by-step breakdown of any part of this process?

1. How is the logarithmic property applied in similar problems?
2. What if the base had been different?
3. How can we confirm both solutions are valid?
4. How would rounding affect this answer?
5. Why is it important to use natural logarithms here?

Tip: Double-check calculations by plugging approximate values back into the original equation to verify accuracy.