How can it be shown that $x^x = a$ , has no real solutions when $a < (1/e)^{(1/e)}$

Question

Using calculus to find the minima:

$$y(x) = x^x$$

$$ln(y) = x*ln(x)$$

$$(1/y)*\frac{dy}{dx} = ln(x) + x*\left(\frac{1}{x}\right) = ln(x) + 1$$

$$\frac{dy}{dx} = y*(ln(x) + 1)$$

$$\frac{dy}{dx} = (x^x)*(ln(x) + 1)$$

Though arriving at this next step, one can assume from looking at it graphically, that $x^x$ will never be $0$, thus $(ln(x) + 1) = 0$, however how can it be shown that $(x^x)$ is never $0$, instead of making a bold assumption?

$$0 = (x^x)*(ln(x) + 1)$$

$$ln(x) = -1$$

$$x = exp(-1) = \frac{1}{e}$$

$$y = \left(\frac{1}{e}\right)^{\left(\frac{1}{e}\right)} ~= 0.6922$$

Answer 1

From the definition,

$$x^x=e^{x\log x}>0.$$

An exponential is always positive.

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The case of $x=0$ is debatable and in fact $x^x$ is not really defined at zero. But for this discussion to make sense, we shoud adopt a definition that makes the function continuous and assign the value

$$\lim_{x\to0}x^x=1.$$

Answer 2

I don't think that's a bold assumption. For instance, $x^x=e^{x\ln x}$ is never zero, as $e$ raised to the power of any real number is strictly positive.

Alternatively, there are no strictly positive $a,b$ that makes $a^b$ zero, and restricting our attention to the special case of $a=b$ doesn't change that.

As for exactly what happens at $x=0$, that's a matter of definitions, not a matter of calculation. I think $0^0=1$ is most sensible, but others may disagree.

The rest of your proof seems good.

Answer 3

Show that $x^x = a$ has no real solutions when $a < \left(\frac1e\right)^{\frac1e}$.

Enough to visualize that

• $\displaystyle\lim_{x\to 0}x^x=\displaystyle\lim_{x\to 0}e^{x\ln x}=e^{\displaystyle\lim_{x\to 0}x\ln x}=e^{\displaystyle\lim_{x\to 0}\frac{\ln x}{\frac1x}}=e^{\displaystyle\lim_{x\to 0}\frac{\frac1x}{\frac{-1}{x^2}}}=e^{-\displaystyle\lim_{x\to 0}x}=1$
• $\displaystyle\lim_{x\to +\infty}x^x\to+\infty$
• $f(\frac1e)=\left(\frac1e\right)^{\frac1e}$
• $$f^\prime(x)=x^x(1+\ln x)= \begin{cases}<0&x\in(0,\frac1e)\\0&x=\frac1e\\>0&x\in(\frac1e,+\infty) \end{cases}$$ So, $f(x)=x^x$ is tangential to the line $y=\left(\frac1e\right)^{\frac1e}$.