The solution set of the equation $x^{10}=10^x$

Question

The solution set of the equation $x^{10}=10^x$ consists of
$(A)$one negative and one positive number
$(B)$one negative and two positive numbers
$(C)$two positive numbers
$(D)$two negative and one positive number

I think the solution set has one negative and a positive number but the answer given is $(B)$ in my book.I did not understand how.May be that is not correct.I want to verify whether my answer is right or not.If not what is the proper way to solve it.

Answer 1

Hard to argue with the graph in the other. But here's one way to look at it:

$0^{10} = 0$, $10^0 = 1$ so $10^x > x^{10}$ at x = 0. $2^{10} = 1024 > 10^2 = 100$ so $10^x < x^{10}$ as x = 2. So there is an x where $x^{10} = 10^x$ for some x; $0 < x < 2$.

$10^{10} = 10^{10}$ so $x^{10} = 10^x$ at $x = 10$.

So that's two positive points where they are equal.

For $x < 0$, $0 < 10^x \le 1$. But $(-1)^10 = 1 > 10^{-1}$. So there is an $x; -1 < x < 0$ where $10^x = x^{10}$.

So of the options B) is the only one that fits.

But to prove there aren't any other solutions. Note: $x^{10}$ and $\frac {d x^{10}}{dx} = 10x^9$ and $10^x$ and $\frac {d 10^x}{d x} = \ln 10 *10^x$ are monotonically increasing on positive x. As $x^{10} < 10^x$ at zero yet $x^{10} = 10^x$ at the first positive solution, this must mean $\frac {d x^{10}}{dx} > \frac {d 10^x}{d x}$ at the first solution. For this x to the second solution $x^{10} > 10^x$ and at the second solution $\frac {d x^{10}}{dx} < \frac {d 10^x}{d x}$ As the second derivitives are also increasing on positive x, $10^x > x^{10}$ for all x greater than the second solution and there'll never be any more positive solutions.

By similar argument there is at most one solution for x < 0. And then only because 10 is even.

In general, for $x^b$ and $b^x$; $b > 1$. The will be two positive solutions $1< x_1 < e < x_2$ if $b \ne e$ and one of the $x_i = b$. If $b = e$ there is one solution at $b = e$. There will be one negative solution if b is even and none if b is odd.

Answer 2

One little picture says more than a long speech!

The three roots appear on the next graph. The approximate values are obtained thanks to numerical calculus.

Instead of $x^{10}=10^x$ consider $x^2=10^{x/5}$ (in order to avoid big numbers and reduce the size of the graph) Keeping an even exponent $2$ preserves the negative range of $x$.

NOTE : (outside the limited scope of the question):

In addition to the trivial root $x=a$, the analytic solution of the equation $x^a=a^x$ is expressed thanks to the LambertW function : $$x=-\frac{a}{\ln(a)}W\left(- \frac{\ln(a)}{a} \right)$$ In case $a=10$ : $$x=-\frac{10}{\ln(10)}W\left(- \frac{\ln(10)}{10} \right)$$ The LambertW function is multivaluated : The two real values in the present case are written on the figure.

Answer 3

Consider the values of the functions $x^{10}$ and $10^x$ at $-1, 0, 2,$ and $100$ (okay, don't actually compute $100^{10}$ vs. $10^{100}$, but figure out which one is bigger - note that $100=10^2$ . . .). This shows that the functions' graphs must intersect at at least three points, one between $-1$ and $0$ (so negative) and the others between $0$ and $2$ and between $2$ and $100$ (so positive).

This rules out all the other possibilities.

Answer 4

Take the $10$th root: $$\left\lvert x\right\rvert=\left(\sqrt[10]{10}\right)^x$$ Write with $e$: $$\left\lvert x\right\rvert=e^{\frac{1}{10}\ln(10)x}$$ Put on the same side: $$\left\lvert x\right\rvert e^{-\frac{1}{10}\ln(10)x}=1$$ Make coefficient resemble exponent: $$\left(-\frac{1}{10}\ln(10)\left\lvert x\right\rvert\right)e^{-\frac{1}{10}\ln(10)x}=-\frac{1}{10}\ln(10)$$ Cases, based on $x$ positive or negative: $$\left(-\frac{1}{10}\ln(10)x\right)e^{-\frac{1}{10}\ln(10)x}=-\frac{1}{10}\ln(10)\qquad \left(-\frac{1}{10}\ln(10)x\right)e^{-\frac{1}{10}\ln(10)x}=\frac{1}{10}\ln(10)$$

Apply $W$: $$-\frac{1}{10}\ln(10)x=W\mathopen{}\left(-\frac{1}{10}\ln(10)\right)\mathclose{}\qquad -\frac{1}{10}\ln(10)x=W\mathopen{}\left(\frac{1}{10}\ln(10)\right)\mathclose{}$$ $$x=-\frac{10}{\ln(10)}W\mathopen{}\left(-\frac{1}{10}\ln(10)\right)\mathclose{}\qquad x=-\frac{10}{\ln(10)}W\mathopen{}\left(\frac{1}{10}\ln(10)\right)\mathclose{}$$

$W$ outputs two (negative) values based on which branch you are looking at for the small negative input on the left solution. On the right solution, $W$ only outputs one (positive) value. Accounting for the sign change in the solution, there are two positive solutions and one negative solution.