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

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#1
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 ?
0
7 months ago
#2
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).
1
#3
(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.
0
7 months ago
#4
(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.
0
7 months ago
#5
(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 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 (by series expansion, for example). You then define , where the log function is the inverse of the exponential function.
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