Solve for X in a difficult exponential function

Question

Solve for $X$ when

$3^{x^x}=1000$

By hand please (without evaluating the intersection on the graph). How is it done?

Answer 1

$$3^{(x^x)} = 1000 \to x^x \ln 3 = 3\ln 10 \to \ln(x^x) + \ln(\ln3)=\ln 3 + \ln(\ln 10) \to\\x\ln x = \ln 3 + \ln(\ln 10)-\ln(\ln 3) $$ that is how far i can go without resorting to some numerical approximation. my ti-83 gives me an approximate solution $x = 2.25772568.$

Answer 2

With a calculator, I would start by taking base $3$ logs, getting $$x^x=\log_3 (1000) \approx 6.2877$$ then just try things. $2$ is somewhat too small, $3$ is way too large, so $$2.2^{2.2}\approx 5.67\\2.3^{2.3} \approx 6.80\\2.28^{2.28}\approx 6.548\\2.25^{2.25}\approx 6.200\\2.255^{2.255}\approx 6.257\\2.256^{2.256}\approx 6.268\\2.258^{2.258}\approx 6.291$$and declare victory. Of course you can use the secant method or Newton's if you want.

Answer 3

Clearly x^x is neither integer nor rational because 100 is rational. Besides if x^x were algebraic then 3^(x^x) would be transcendental according to the famous Gelfond–Schneider theorem. Therefore x^x must be transcendental and 4 < x^x <5 because 3^4 < 100 < 3^5 so we have 4 < x^x <5. From x^x log3= 2 it follows x^x = 2/log3 and I leave the problem without graphing or numerical calculator….