The opposite of Exponent in Javascript? - Overclock.net

SgtSpike · 02-19-07

So I'm working on a project in javascript, and I wrote a short function to compute x^y. Now I need to do the opposite, and I'm not sure how. Here's the original function.

Code:

function exp(a,b)
{
t = 1;
 for(x = 1; x <= b; x++)
    {
     t = t * a;
    }
EXP = t;
return(EXP);
}

So if I put in exp(2, 3) (which would be 2^3), I will get 8 as the returned value. Then, in a new function I need to do the opposite. Given 8 and 2, I need to figure out what y would be (in this case 3). How can I do this?

HrnyGoat · 02-19-07

So you would be looking for the root? You could just put in a fraction as the exponent value, like x^1/2 to get the square root.

SgtSpike · 02-19-07

Not exactly. I would need to find the opposite of what you are talking about. Instead of finding x, I already know it. I would need to find the 1/2.

Janus67 · 02-19-07

I believe you are looking to do a logarithm. you may need to find a .jar file to help you do that.

edit:

I haven't coded in java for a year so it may be off by a little bit, but I think this is what you were looking for.

that or upon thinking about it, just take the value as it is.

y=2
z=8

for(x=1; x<<put in a number here that you would want it to stop incase it isn't exact>;x++)
{
if(y==z)
break;
else
<call function for your exponential function and put in x,y and return that value>
}

C-bro · 02-20-07

Well the actually Java function uses log(x) which is the natural logarithm (normally denoted ln(x)) it can be found in the java.lang.Math class.

For your project, the inverse of multiplication is division, so why not do:

Code:

public static int log(double a, int b){ // computes log of a, base b 
int exponent=0;
double remainder=a;

    do{
        remainder /= b;
        exponent++;
    }while(remainder > 1);
return exponent;
}

What this does is figure out the value exponent such that

a = b^exponent

Note: My method will only work when a is a perfect power of b. In other words, exponent must be an integer. You're going to have to do some working if you want to integrate decimal exponents.

In other words
log(8,2) = 3
log(16,2) = 4

But for a number where the exponent is not an integer, the value gets rounded up.

log(9,2) = 4 rather than 3.1699

You must define your accuracy and consider if this method will suffice for your program constraints.

SgtSpike · 02-20-07

Hmmm... Great starts Janus and C-Bro.

C-Bro - I like your method, it's exactly what I need except I do need decimal-level precision to 8 or 9 places... Any other thoughts on how to do that? Could I have a +1 loop to find the closest integer digit, and then a +0.1 loop to find the closest first decimal digit, and then a +0.01 loop to find the next decimal digit, etc. Would that work? What would it look like?

rabidgnome229 · 03-01-07

Dunno if you still need this - but if so

Code:

public static double log(double a, int b, int places){ // computes log of a, base b to <places> decimal places
    double exponent=0;
    double increment = 1.;
    double remainder=a/b;
 
    for(int i = 0; i < places; i++)
    {
        while(remainder>1)
        {
            exponent += increment;
            remainder /= b;
        }
    
        increment /= 10;
    }
    return exponent;
}

You may want to test it for bugs but that should work

C-bro · 03-07-07

I'm almost sure you won't need this now, but I used this method on test today for implementing the ln() function. It was said that we had an exp function available to us.

The basis behind the code is the Newton-Raphson method of approximation. You're basically finding the zeros for the inverse function to solve for the answer. Say we want to find ln a.

x = ln a
exp(x) = a
exp(x) - a = 0

Hopefully none of that is foreign so far.
You can then take the NR method which is:

Xi+1 = Xi - f(Xi)/f'(Xi)

I used a fixed guess of X0 = 1, and it produces accurate results to the maximum I can display in 8 iterations.

This finds ln a. I tried it for around 10 or 15 different values and they were correct to 14 decimal places (the most I could view). Also note this was done in base e, but can be modified to work for any base.

Code:

int guess = 1;
for (int i=0;i<7;i++)
    guess=guess - (exp(guess)-a)/exp(guess);
answer = guess;

SgtSpike · 03-08-07

Actually, thanks for the great replies! I can still use this definitely. I had just put the project on hold, but I'll probably resume it here shortly. I'll implement those algorithms and see what I come up with. Thanks and Rep+!

C-bro · 03-08-07

I have found a flaw in my method. It works perfectly for the natural logarithm like I posted there, but it falls apart when working with other bases. Why is that?

For a base b, the method would go as follows:

logb a = x
b^x = a
b^x - a = 0

However, when you try and differentiate the b^x term you get stuck with
d/dx(b^x) = b^x * 1 * ln b

The reason it worked so well with base e is because the derivative of e^x = e^x. If you want to use the Newton-Raphson approximation, you'll have to include both a ln(a) and log(a,b) function. I've been working with these methods and it really doesn't seem to be a great algorithm for log with any base. In fact when the numbers get to high, it downright crashes. log(2048,2) produces NaN when it should be 11. For large difference between base and product, a huge number of iterations are also needed. I'd say abandon this approach. It's smart on paper but doesn't seem practical. Here's what I finished with when I gave up. Every thing "works" but only up to a limit.

Code:

    public static double log(double a, double b){
        
        double guess = 1;
        for (int i=0;i<1000;i++)
            guess=guess - (pow(b,guess)-a)/(ln(b)*pow(b,guess));
        return guess;
    }
    
        public static double ln(double a){
        
        double guess = 1;
        for (int i=0;i<7;i++)
            guess=guess - (java.lang.Math.exp(guess)-a)/java.lang.Math.exp(guess);
        return guess;
    }

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The opposite of Exponent in Javascript?

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