Solutions to equation $x^{\frac2 7}=16$

Question

My son had this question on his college algebra homework. Solve $x^\frac{2}{7}=16$.

The question doesn't specify the domain, but they have learned about complex numbers.

It's an online homework assignment, and it tells him if his answer is correct and will let him change to a different answer. The online system would not accept the answer 16384, but would only accept the answer {16384, -16384}.

R says $(-16384)^\frac{2}{7}$ is not a number (NaN).

Mathematica says $(-16384)^\frac{2}{7}=16 (-1)^\frac{2}{7}$, which begs the question: What is $(-1)^\frac{2}{7}$?

I understand that $((-16384)^2)^\frac{1}{7}=16$, but $((-16384)^\frac{1}{7})^2$ is again not a number according to R and is $16 (-1)^\frac{2}{7}$ according to Mathematica.

Finally, using Solve or Reduce in Mathematica, it says there is only one solution $x=16384$.

Answer 1

There are two different conventions at work here which lead to different answers.

First their is the basic algebra convention that $x^{\frac{a}{b}}$ for $\frac{a}{b}$ a fully reduced fraction should be interpreted as $(x^a)^{\frac{1}{b}}$. If you interpret it this way, then indeed $x=-16384$ and $x=16384$ are the two solutions to the equation $x^{\frac{2}{7}}=16$.

Second there is a general interpretation of $x^y$ for $x \in \mathbb{C} \setminus \{0\}$ and $y \in \mathbb{R}$. Write $x=r\cdot e^{i\varphi}$ in polar coordinates, with $ r > 0$ and $\varphi \in [0, 2\pi)$, then $x^y = (r\cdot e^{i\varphi})^y = r^y \cdot e^{iy\varphi}$. This is also a complex number in polar coordinates if you interpret the $y\varphi$ modulo $2\pi$. With this interpretation there is only a single complex solution to $x^{\frac{2}{7}}=16$, namely $x=16384$, but $x=-16384$ is not a solution.

Answer 2

By the usual conventions used in College Algebra, $x^{a/b} = (x^a)^{1/b}$ if $x$ is real and $a/b$ is a rational number in lowest terms with $a$ even.

Answer 3

$$x^{\frac27} =16$$

$$(x^{\frac27})^{\frac72} =x =16^{\frac72}=\pm 4^7=\pm16384 $$

That is because we can multiply the repeated exponents in this case and the $\pm$ is for the square root of 16 admitting two roots of opposite sign. As the exponent 7 is odd, negative sign carries through.

Answer 4

In analysis and calculus $b^x:= e^{x\ln b}$ and if $b \le 0$ you will get $NaN$ and that is probably what is going on with R.

But in other context if $r\in \mathbb Q$ then $b^r=b^{\frac nm}:= (b^n)^{\frac 1m}$ where $r=\frac nm$ expressed in lowest terms. In this case if $m$ is even and $b$ is negative we are out of luck but otherwise its just fine. So $(-1)^{\frac 27} = ((-1)^2)^{\frac 17}=1^{\frac 17} = 1$. And that's probably what's going on everywhere else.

But...

And... just to go nuts... in complex analysis we convert a complex number $a+bi; a,b\in\mathbb R$ to form $r(\cos \theta + i\sin \theta) = re^{\theta i}$ were $r= |a+bi|$ and $\theta = arg(a+bi)$. So in that case:

$-1 = 1e^{(2\pi k +\pi) i}$ and $(-1)^{\frac 27}$ will be the seven complex numbers

$e^{\frac {4\pi k + 2\pi}7i} = \cos \frac {4\pi k + 2\pi}7 + i \sin \frac {4\pi k + 2\pi}7$ for $k = 0,$ to $6$. Of these $7$ complex numbers the one where $k=3$ yields

$\cos\frac {12\pi + 2\pi}7 + i\sin \frac {12\pi + 2\pi}7 = \cos 2\pi + i \sin 2\pi = 1+ i\cdot 0 = 1$.

Which, lo and behold, is exactly what we wanted it to be.

Answer 5

R returns NaN for complex numbers unless you specify as.complex() or use functions designed for complex numbers. You can use polyroot to find all of the roots of a polynomial. The function takes the a vector of polynomial coefficients in ascending order as the argument. In this case, you can solve $16^7-x^2 = 0$:

polyroot(c(16^7, 0, -1))
[1]  16384+0i -16384+0i

Or $x = (-1)^{2/7} \longrightarrow x^7-1=0$:

> polyroot(c(-1, 0, 0, 0, 0, 0, 0, 1))
[1]  0.6234898+0.7818315i -0.9009689+0.4338837i -0.2225209-0.9749279i  0.6234898-0.7818315i
[5] -0.2225209+0.9749279i -0.9009689-0.4338837i  1.0000000+0.0000000i

Answer 6

If you take the absolute value of both sides, you get:

$$|x|^{2/7} = 16$$ $$|x|^2 = 16^7 = 2^{28}$$ $$|x| = 2^{14} = 16384$$

But what if $x \in \mathbb{C}$? Recall that according to DeMoivre's formula, that if $x$ is expressed in polar form as $x = r(\cos\theta + i\sin\theta)$, with $r = |x| \ge 0$, then:

$$x^n = r^n \left(\cos(n\theta) + i\sin(n\theta) \right)$$

Plugging in $r = 16384$ and $n = \frac{2}{7}$ gives:

$$x^{2/7} = 16384^{2/7} \left(\cos(\frac{2\theta}{7}) + i\sin(\frac{2\theta}{7}) \right) = 16$$ $$16 \left(\cos(\frac{2\theta}{7}) + i\sin(\frac{2\theta}{7}) \right) = 16$$ $$\cos(\frac{2\theta}{7}) + i\sin(\frac{2\theta}{7}) = 1$$ $$\frac{2\theta}{7} = 2\pi k, k \in\mathbb{Z}$$ $$\theta = 7\pi k, k \in\mathbb{Z}$$

Thus:

$$x \in \{ 16384\left(\cos(7\pi k) + i \sin(7\pi k) \right) : k \in \mathbb{Z} \}$$

But $\sin$ is 0 for any integer multiple of $\pi$, and $\cos(7\pi k)$ must be either $1$ (if $k$ is even, and thus $7k$ is even), or $-1$ (if $k$ is odd, and thus $7k$ is odd). Thus,

$$x = \pm 16384$$