Why does x raised to an odd power result in a unique solution?

Question

I don't think I'm phrasing this correctly but anyway, say we are looking for solutions to x for $x^9=1/2$,

why is there only one unique solution as given by $x=(1/2)^{1/9}$

when by raising x to an even value gives two solutions as $\pm$ solution

I think I'm just missing a key concept so any help would be appreciated.

Answer 1

The simple answer is there is not only one solution, but only one real solution. For example, the equation $x^3=1$ has the solutions $x=1,e^{\frac{2i\pi}{3}},$ and $e^{\frac{4i\pi}{3}}$.

Answer 2

Look at the graph of $y = x^9$ (or $y = x^3$, or any other such thing). (Probably best to start with $y = x^1$ for simplicity.)

You'll see that a horizontal line intersects the graph only in a single point. So if you know $x^9$, there can only be one corresponding $x$-value.

On the other hand, graphs of things like $y = x^2$ look like "bowls", and for each non-negative $y$-value, there are two corresponding $x$ values (which are negatives of one another.)

Answer 3

$x^9=1/2$ has exactly 9 solutions by the Fundamental Theorem of Algebra, however the principal root is $2^{1/9}$

Raising $x$ to the even power, say $x^2=1$ would make it so there can be two solutions $\pm1$, since $x^2-1=0$ evaluates to $(x+1)(x-1)=0$, so there are positive AND negative roots. Another way to think about this is if you plug $-1$ into $x^2=1$ for $x$, you have $(-1)*(-1)=1$, which is correct.

On the other hand, if you raise $x$ to an odd power, say $x^3=1$, you obtain only 1 real solution, which is $1$. $-1$ does not work because $(-1)*(-1)*(-1)=-1$, not $1.$

Answer 4

Raising to an even power discards the sign (the number becomes positive). Raising to an odd power retains the sign.

To see that, consider that if $x$ is negative, raising $x$ to an even power means multiplying negative numbers an even number of times. You can pair the numbers and each pair multiplies to a positive number. Hence the product is positive.

So, because the sign is discarded, the solver is free to use the positive or the negative number.

Raising $x$ to an odd power, say, $a$, is equivalent to raising $x$ to the $(a-1)$th power then multiplying the result by $x$. $a-1$ is even, so $x^{a-1}$ is positive. So, the sign of $x^a$ is the sign of $x$.

So, because the sign is retained, the solver has to use the number with the same sign as the right-hand sife.

Answer 5

Another way of hand-waving the same result:

If the largest power of $x$ in an equation is odd, then the sign of the expression must change as $x$ goes from a very negative value to a very positive value. Somewhere in between those two values, for at least one value of $x$, the expression must be zero.

OTOH, if the largest power of $x$ is even, then the sign of the expression will be the same for very negative values of $x$ and for very positive values of $x$. You cannot say with certainty whether the expression will ever be zero in this interval...