Please help with C++ power to

[lucky78]

ok I have been stuck on this for quite a while now

even when I try using math::pow it doesn't seem to work

well let me explain what i needed to do

i needed to create a simple interest and a compound interest calculator

i did the first part of it just fine

like so :

// Declare variables
double P, R, T, sum;

// Read in data
Double::TryParse(txtNum1->Text,P);
Double::TryParse(txtNum2->Text,R);
Double::TryParse(txtNum3->Text,T);

// Proccess data
sum = P * (1 + (T *(R/100)));

// Display results
txtBx4->Text = sum.ToString();


but when it comes to do the compound interest i need to work with powers of
and that seems to be the problem

heres what i need

// Declare variables
double P, R, T, sum;

// Read in data
Double::TryParse(txtNum1->Text,P);
Double::TryParse(txtNum2->Text,R);
Double::TryParse(txtNum3->Text,T);

// Proccess data
sum = P * (1 + (R/100)^to the power T

// Display results
txtBx5->Text = sum.ToString();





any help would be greatly appreciated thanks


and by the way am using visual studio if that helps any

[buffbill]

say x=1+(R/100) then (1 + (R/100)^to the power T ==pow(x,T) //pow returns a double.
You will need a header file <cmath>.

[lucky78]

I tried a header <math.h> but got a whole bunch of errors

I'm quite new to c++ can you be a little more specific please

[Serious Sam]

What errors did you get? Not sure if this is what you want, but working with the GNU compiler the following code works:

#include<iostream>
#include<math.h>
using namespace std;

int main() {
double a = 2.0;
double b = 3.0;

double c;

c = pow(2.0,3.0);
cout<<c; //prints 8 to the standard output
}

So in your case the line: sum = P * (1 + (R/100^to the power T))

would be: sum = P * (1 + pow((R/100),T));

if you want T to be the exponent and (R/100) to be the base. Hope that helps.

EDIT: "using namespace std;" has nothing to do with the pow-function, it's just to make the name cout for standard output visible.