OT: Interesting math problem...

[Kamog]

If you're allowed to use logarithms:

sqrt (x) = e^(0.5 * ln(x))

[narf poit chez BOOM]

Thanks Jack. But, still not an equation.

Kamog, what?

[Jack Simth]

Kamog said:
If you're allowed to use logarithms:

sqrt (x) = e^(0.5 * ln(x))

Kamog, I don't mean to be picky, but if you have decimal exponents available to do the e^ portion, you don't need logarithms at all:

sqrt (x) = x^(0.5)

- it's a basic identity.

[Kamog]

Jack Simth said:
Kamog, I don't mean to be picky, but if you have decimal exponents available to do the e^ portion, you don't need logarithms at all:

sqrt (x) = x^(0.5)

- it's a basic identity.

Hmm, that's a good point.

Yeah, I know that identity. Actually I started off with it to get the equation:
y = x^0.5
ln(y) = ln(x^0.5)
ln(y) = 0.5*ln(x)
y = e^(0.5*ln(x))

I see what you mean, why use logarithms if you have decimal exponents available.

[narf poit chez BOOM]

So, how do you calculate 64^0.5 without a calculator?