How do you answer questions where a polynomial equation is set equal to an exponential? The question I found was:
How many solutions does $2^x = x^{12}$ have?
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Sorry, I made a mistake. The question was actually $2^x = x^{12}$. In any case, the answer manual says there are 3 solutions. – Fernandeez nuts Jan 28 '24 at 12:33
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1$3$ solutions where? In integers? – Dietrich Burde Jan 28 '24 at 12:47
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$(\mathbb{R})^{2}$ – Fernandeez nuts Jan 28 '24 at 12:48
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Why $(\Bbb R)^2$? I only see one variable $x$. Do you mean in $\Bbb R$? See also this post. And, actually, the very same equation has been solved at art of problem solving. So $x=-0.946780330394$ or $x=1.06334683067$ or $x=74.6693255347$. – Dietrich Burde Jan 28 '24 at 12:49
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To find all the solutions, you would need to employ the Lambert $W$ function. However, it is easier to find the real solutions. Let $x>0$, and then take logarithm on both sides $$\frac{x}{\ln x} = \frac{12}{\ln 2} \approx 17.3$$ Now define $f(x) := \frac{x}{\ln x}$. Note that $f(x)$ is continuous and differentiable on $\mathbb R^+ / \{1\}$. Take its derivative $f'(x) = \frac{\ln x -1}{\ln^2(x)}$. From the derivative, we can conclude that $f(x)$ is decreasing in $(0,1)\cup (1,e)$ and increasing on $(e, \infty)$. Now since $f(e)=e$, and $\lim_{x \to 1^+} f(x) = \infty$, it follows there is exactly one solution in $(1,e)$. Now since $f(x)$ also monotonically increases from $(e, \infty)$, there must also be exactly one more solution in $(e, \infty)$.
Finding if a negative solution exists is also pretty easy. We are essentially looking for number of positive solutions to $\frac{1}{2^y} = y^{12}$ and our negative solution will be the values of $-y$. Again taking the natural logarithm we get $$\frac{-y}{\ln y} = \frac{12}{\ln 2}$$ We see this means $f(y)=-\frac{12}{\ln 2}$. Now since $\lim_{x \to 1^-} f(x) = -\infty$, and $f(0)=0$, and $f(x)$ decreases on $(0,1)$, there must be exactly one solution $y \in (0,1)$. This means our third solution is in $(-1,0)$.
Sahaj
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