Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
You are not logged in.
Post a reply
Topic review (newest first)
The function takes in the number in double and the base in integer such that when the function is called:
When we are trying to find the log of x to the base of b equals "I" then this is the same as:
and this is the same that this code is using.
First of all if the log number is negative, this code gives an error because logs are only defined for Number > 0.
If our number is nonnegative then the code proceeds and we enter a For loop from I = 0 to Number. The first check this For loop does is to see if Base ^ I = Number because if thats the case, it means that "I" is our answer to the log, therefore if thats the case then we just say that LogB=I and exit the function because we just found it.
If this is not the case then the For loop does another check and this check is to see if Base^I is less than the Number and Base^(I+1) is greater than the Number. The loop will continue looping until we find that I that makes Base^I less than Number and Base^(I+1) greater than Number. When we find this I value, this means that the answer to the log is between I and (I+1) and this is because of what we said before.
Remember that when we had a log, it was the same as this:
Base^I = Number
So in the code, if we have:
Base^I < Number
Base^(I+1) > Number
that tells us that the log is greater than I and less (I+1) so our answer is in the interval (I,I+1).
Ok, so now that we know our answer must be between I and (I+1) then the code creates another For loop which runs from 0 to 99999 and what this does is that it keeps adding 0.00001 to Lg (Lg = I) and keeps checking if Lg is our answer. We are bound to find it because we knew that our answer was between I and I+1 so if kept adding 0.00001 to I and checking, add another 0.00001 and check again, ..... , eventually we will find our answer to the log.
can anyone comment this log function code
Function LogB(Number as double,Base as integer) as Double Dim Lg as Single If Number <=0 then Err.Raise 5 , , "Logarithms for zero and negitive numbers are not defined" Exit Function End If For I = 0 to Number If Base ^ I = Number then LogB=I Exit Function ElseIf Base ^ I < Number And Base ^ (I+1) > Number then Lg = I For J= 1 to 99999 Lg = Lg + 0.00001 If Base ^ Lg >= Number Then LogB = Lg Exit Function End If NExt J End IF Next I End Function
I can't understand this how he build log function