23 = 2 * 2 * 2 = 8
Many people want a block like this to make exponents easier to program:
(() ^ ():: operators)
Using the Method in Scratch
There are multiple ways to do exponents in Scratch. Here are two:
|Note:||this method can only be used with whole numbers|
This script will be able to solve exponents in Scratch; however, it cannot calculate decimals:
when flag clicked ask [What is the base number?] and wait set [1st# v] to (answer) ask [To what power?] and wait set [2nd# v] to (answer) set [ans v] to (1) repeat ([abs v] of (2nd#)) set [ans v] to ((ans) * (1st#)) end if <(2nd#) < > then set [ans v] to ((1) / (ans)) // a quick workaround for negative powers end say (join [The answer is: ] (ans)) for (2) secs
How it Works
- The Variable "1st#" is the base number. This is the number that will be multiplied by.
- The Variable "2nd#" determines how many times to repeat the multiplication.
- The variable "ans" is the answer, which is the number 1 multiplied by "1st#", "2nd#" times.
The logarithm method is much faster than the repeat method.
([10 ^ v] of ((power) * ([log v] of ([abs v] of (base)))))
([e ^ v] of ((power) * ([ln v] of ([abs v] of (base)))))
How it works
- The variable "base" is the base and the variable "power" is the power, or the number of times the base is multiplied. The reported value is the answer.
- The math works because n * log m = log mn and 10log mn = mn.
These are all errors, so it is advised to have cases for them:
- The base is 0: Since log 0 is undefined, have an if-else that gives 0 for the answer if the base is 0, as long as the power is positive. The answer 1 should be given for 00 so that it does not give an error then.
- The base is negative: If one wants to be able to do something such as xp where p is an integer, and they want to allow x to be negative, then the solution with logarithms will not work, because log(-1) is undefined (there is no solution to 10x = -1). Ways to solve this are:
- Use the loop multiplication method, and even just use loop multiplication / division whenever the power is an integer.
- Factor out the negative and find the answer for the absolute value of the exponent, then divide 1 by it if the exponent is negative. Also, it's a good idea to use an if block for whether p is odd or even ( if p mod 2 = 0 then p is even).
set [result v] to ([abs v] of (base)) // used to spread out math over two lines set [result v] to ([e^ v] of ((power)*([ln v] of (result)))::operators) if<(base) < > then set [result v] to ((1) / (result)) end
|Note:||Positive rational exponents, e.g. 10½ do work. Thus, this can be used to find N-roots of numbers.|
|This article or section documents a feature not included in the current version of Scratch (2.0). It is only useful from a historical perspective.|
('%n ^ %n' #r #exp:of:)
exp:t1 of:t2 ^ t1 raisedTo: t2