What calculation would I do to calculate 2^ π ? What logic is behind the indicy rule?

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lhh2003
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2^2 = 2*2
2^3 = 2*2*2

2^1.5 = √(3^3) = 3√3 .

Now, intuitively , the previous calculation would be 2 multiplied by itself 1.5 times which is kind of weird to compose an explicit calculation for other than what I have learnt parrot fashion from year 9 of the indicy rule to answer that question. This only works for rational numbers.

To find 2^ π, I guess I'd just have to round π and apply the indicy rule to a fraction a/b where a and b are rational numbers.

But I can't see the logic behind the indicy rule ; what was the inventor's logic ?
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Justvisited
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If a^n = b, then a is the nth root of b. Then simply apply the addition law of indices to realise that a = b^(1/n).
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lhh2003
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(Original post by Justvisited)
If a^n = b, then a is the nth root of b. Then simply apply the addition law of indices to realise that a = b^(1/n).
Thanks.
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mqb2766
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(Original post by lhh2003)
2^2 = 2*2
2^3 = 2*2*2

2^1.5 = √(3^3) = 3√3 .
2^1.5 = √(2^3) = 2√2.
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Gregorius
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(Original post by lhh2003)
To find 2^ π, I guess I'd just have to round π and apply the indicy rule to a fraction a/b where a and b are rational numbers.
The way you're suggesting ties in with one of the standard ways of defining the power operation for irrational powers: demand that the function x \rightarrow b^x be continuous, and then approximate the irrational exponent with a sequence of rational exponents, and take the limit.

The way it's more usually done is to first of all define the exponential function e^x (by series expansion, for example). You then define b^x = e^{x \log{b}}, where the log function is the inverse of the exponential function.
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