1. **State the problem:** We need to find the temperature $T$ given the equation $$30000 = 101325 \left(\frac{T}{288.16}\right)^{-\frac{9.81}{-6.5 \times 10^{-3}} \times 287.08}.$$\n\n2. **Identify the formula and variables:** This equation resembles the barometric formula relating pressure and temperature in the atmosphere. Here, 30000 and 101325 are pressures (in consistent units), $T$ is the temperature to find, and constants are given.\n\n3. **Simplify the exponent:** Calculate the exponent first. The exponent is $$-\frac{9.81}{-6.5 \times 10^{-3}} \times 287.08.$$\nCalculate denominator: $$-6.5 \times 10^{-3} = -0.0065.$$\nThen, $$-\frac{9.81}{-0.0065} = \frac{9.81}{0.0065} \approx 1509.23.$$\nMultiply by 287.08: $$1509.23 \times 287.08 \approx 433,334.5.$$\nSo the exponent is approximately $$-433,334.5.$$\n\n4. **Rewrite the equation:** $$30000 = 101325 \left(\frac{T}{288.16}\right)^{-433,334.5}.$$\n\n5. **Isolate the power term:** Divide both sides by 101325: $$\frac{30000}{101325} = \left(\frac{T}{288.16}\right)^{-433,334.5}.$$\nCalculate left side: $$\frac{30000}{101325} \approx 0.296.$$$\n\n6. **Take logarithms:** Let $$x = \frac{T}{288.16}.$$ Then $$0.296 = x^{-433,334.5} = \frac{1}{x^{433,334.5}}.$$\nInvert both sides: $$\frac{1}{0.296} = x^{433,334.5} \Rightarrow 3.378 = x^{433,334.5}.$$\n\n7. **Solve for $x$:** Take the natural logarithm: $$\ln(3.378) = 433,334.5 \ln(x).$$\nCalculate $$\ln(3.378) \approx 1.215.$$\nSo, $$\ln(x) = \frac{1.215}{433,334.5} \approx 2.804 \times 10^{-6}.$$\n\n8. **Find $x$:** $$x = e^{2.804 \times 10^{-6}} \approx 1 + 2.804 \times 10^{-6} = 1.0000028.$$\n\n9. **Calculate $T$:** $$T = x \times 288.16 \approx 1.0000028 \times 288.16 = 288.16 + 0.0008 \approx 288.16.$$\n\n**Final answer:** The temperature $T$ is approximately $$\boxed{288.16}.$$