How would I do the equation
$b^{n}$
Or
$b^{-n}$
Where $b$ is the base, and $n$ in a negative/decimal?
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2 Answers
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4
You follow the rule $$b^{-n} = \frac{1}{b^n}$$ Proving the rule is pretty simple ($b\neq 0$): $$b^n \cdot b^{-n} = b^{n-n} = b^0 = 1 \\ \implies b^n \cdot b^{-n} = 1 \implies b^{-n} = \frac{1}{b^n}$$
Edit: for decimals, it really depends on what the decimal is. For example $4^{0.5}$ is simple because $0.5=1/2$ thus $4^{1/2}=\sqrt{4}=2$. But if you had something tougher like $4^{1.5}$ then the method is similar: $4^{1+0.5} = 4^1 \cdot 4^{1/2} = 4\cdot 2 =8$ But when you make it tougher, like $4^{1.234}$ then it gets tedious to split it up the way we did and write them as fractions and so on. Sometimes you end up with things like $4^{1/3}$ which is the cube root of 4. You can only approximate such things. So unless it's a simple one, one would usually use a calculator.
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$\begingroup$ Could you actually do the 1.234 example? I should understand it at that point. $\endgroup$ Oct 3, 2014 at 13:28
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$\begingroup$ I found another way of doing it! You can convert it to a fraction, this it's super easy. $\endgroup$ Nov 10, 2014 at 4:09
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For the decimal case you have two cases; one if $n \in \mathbb{Q}$ where you can apply $b^{\frac{m}{n}} =\sqrt[n]{b^m}$
If $n \in \mathbb{R} - \mathbb{Q}$ then you have to approximate. Either let a CAS do it or write a simple approximation as a fraction(which will undoubtedly end up hard to calculate by hand).