What do you think about the method of using the logarithm to extract a root?
For example, for the square root of 9 10^(1/2*lg(9)).
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6.94dm³/sx 2500N/dm² | : 0.55 is 340120 Nm/s, right?
6.94dm³/sx 2500N/dm² | : 0.55 is 340120 Nm/s, right?
10^(1⁄2 × log(9)
Think that this should only serve to illustrate. For numerical root pulling, there are simpler procedures.
I find it very good that you understood me despite my mistake, whose correction I requested.
Now he’s corrected.
I don’t think it’s the square root, that’s different.
For roots with any exponents you will probably have to use logarithms. You need a logarithm panel for the manual calculation. Calculators (and also other computers) can calculate logarithms themselves, which have no problem with this.
Nix, the Heron method converges quickly and only needs basic calculations.