# Get the number, knowing the power and result of raising.

If I have a number and a power of this number. How to get the number that raised to the known power will give the known number?

$$x^3 = 90$$

How to get $x$?

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You get the cube root... – J. M. Apr 28 '12 at 7:58

$x^{3(\frac13)}=x=90^\frac13$ or $\sqrt[3]x=x=\sqrt[3]90$

If you also wish to find the complex roots to this problem, rearrange the equation so that $x^3-90=0$.

$x=90^\frac13$, hence you know that $f(90^\frac13)=0$, so $(x-90^\frac13)$ is a factor.

Divide $x^3-90=0$ by $(x-90^\frac13)$, using long or synthetic division.

Find the complex roots of this quadratic you have just obtained, $x^2+\frac{2169}{484}x+\frac{4840}{241}=0$, using the quadratic formula.

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First: there may be two such numbers. Think of $x^2=1$. Second: actually, you are simply defining, if it exists, the inverse function of a power function. Rigorously you'd need some work with real numbers and the axiom of continuity.

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In this case, a number can have three cube roots, not just two... – J. M. Apr 28 '12 at 8:12
Well, I guess the OP is working in $\mathbb{R}$. The map $x \mapsto x^3$ is invertible from $\mathbb{R}$ to $\mathbb{R}$. Reading the question, I understand that the OP is studying equations like $x^n=q$, and not $P(x)=q$ where $P$ is some polynomial. – Siminore Apr 28 '12 at 8:30

Use your knowledge of manipulations of exponents. Recall: $x^{a+b} = x^a \centerdot x^b$ and $(x^a)^b=x^{a \centerdot b}$ for real numbers $x,a,b$. Thus, we can obtain $x$ from $x^3$ by simply raising the latter to the 1/3-power. However, as every elementary algebra student is drilled with, what we do to one side, we must do to the other, so $(x^3)^{1/3}=x^{3 \centerdot 1/3}=x=90^{1/3}$. This is equivalently the cube root of 90.

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