set [a v] to (answer)
repeat (a)
set [b v] to ((b) * (a))
end

15-MinuteGaming wrote:

Support. They have logarithms but no exponents.

Script bars (result)
Set result to base
Repeat ((exponent)-(1))
Set result to result * result
End
Report result

noogai34 wrote:

No support. In Snap! the workaround is…

Script bars (result)
Set result to base
Repeat ((exponent)-(1))
Set result to result * result
End
Report result

result
Yes I was l a a y

(x) ^ (y) :: custom
do something with (result) :: custom

DarthVader4Life wrote:

noogai34 wrote:

No support. In Snap! the workaround is…

Script bars (result)
Set result to base
Repeat ((exponent)-(1))
Set result to result * result
End
Report result

result
Yes I was l a a y

this is scratch

define set answer to (base)^(exponent)
set [result v] to (base)
if <(exponent)=(0)> then 
set [result v] to [1]
else
if <(exponent)>(0)> then 
set [counter v] to [1]
repeat until <(counter)=(exponent)>
set [result v] to ((result)*(base))
change [counter v] by (1)
end
else
set [counter v] to (-1)
repeat until<(counter)=(exponent)>
set [result v] to ((result)*(base))
change [counter v] by (-1)
end
set [answer v] to (join [1/] (result))
end
end

redfireball256 wrote:

...

([e ^ v] of (9))

([10 ^ v] of (9))

scubajerry wrote:

DadOfMrLog wrote:

firedrake969_test wrote:

The thing is that you can't do negative numbers with logs.

OK, so here's my definitive x^y custom block…

define result = (x) ^ (y)
if <(y) = [0]> then
  set [result v] to [1] // yes, we also include 0^0=1  :O
else
  if <(x) = [0]> then
    set [result v] to [0]  // 0^y=0 for any y except zero (see above)
  else
    set [result v] to ([e^ v] of ((y)*([ln v] of ([abs v] of (x)))) // no negative x for now
    if <(x) < [0]> then  // now deal with power of negative number
      if <(round(y)) = (y)> then // we can do integer powers of negative numbers
        if <((y) mod (2)) = [1]> then
          set [result v] to ((0) - (result)) // odd powers will be negative
        end
      else
        set [result v] to [NaN] // but let's not go there at this stage...
      end
    end
    if <<(y) > [0]> and <<(round(x)) = (x)> and <(round(y)) = (y)>>> then
      set [result v] to (round (result)) // ensure we get exactly an integer if both x & y were ints
    end
  end
end

That should cover everything but non-integer powers of negative numbers (which gets a bit hairy…), and it makes sure that integer raised to integer gives exactly an integer (in case you assume you'll get an integer, perhaps because you check for equality with an integer at some point).

Hope I've got that all right!

Have you put this is a project?

360Genius wrote:

yes please! I SUPPORT!

Nambaseking01 wrote:

360Genius wrote:

yes please! I SUPPORT!

Please explain why you support - the Scratch Team unfortunately doesn't implement suggestions based off of the amount of supports but the reason for the support.

—

Anyways, support. This is an important math formula (and way more commonly used than all those trigonometry formulas in the operators section) that many Scratchers would like to have, and the workaround isn't really that… simple. Additionally, I think everyone can agree it's very fun to play around with high numbers in programming, so this would increase a lot of fun (at least for me personally).

(() ^ () ::operators)