How do you use e to power of in this case?

[Dkob1]

so I have this, I'm trying to use the formula B=Pe^(rt) , P is dollars deposited in an account for t years with the interest rate of r. How do I use e to the power of something?? I don't really understand how to use it. thx

#include <iostream> 
#include <cmath> 
#include <iomanip> 
using namespace std;

int main()
{
	//declare identifiers 
	float dollars;
    float rate;
	float time;
    float total; 

    //initialize identifiers
     total= (dollars*pow(exp(x), (rate*time))); 

	cout << fixed << showpoint << setprecision(2);

    //Enter values
	cout << "Enter the money you want to deposit now(P): " << dollars;
	cout << "Enter the interest rate: " << rate;
    cout << "Enter duration in year(t): " << time;
    cout << "   Your balance after 5 years is " << total; 

	cout << endl; 

	//terminate program
	system("pause");
	return 0;
}

[MiiNiPaa]

by e do you mean Euler number? If so, it is easy to get by using exp(1.0)

[Dkob1]

How would I use it to say e^(rate * time) ? Ya thats what I meant.

[Dkob1]

I'm trying to do dollars* e ^(rate * time) to find the new balance.

[MiiNiPaa]

std::pow(e, rate*time)

[Dkob1]

I think I tried that too. Then what would I make e ?

[MiiNiPaa]

What is e exactly in your code?

MiiNiPaa wrote:

by e do you mean Euler number? If so, it is easy to get by using exp(1.0)

[Dkob1]

Yes, but I tried that and it's not working.

[MiiNiPaa]

I tried that and it's not working.

Then you made a mistake.

What exactly is not working? How is it "not working"? What you thinnk should happen?

http://coliru.stacked-crooked.com/a/e32e31b7e4e4b969

[Dkob1]

But instead of numbers I have rate and time though, how do I incorporate that?

[MiiNiPaa]

MiiNiPaa wrote:

std::pow(e, rate*time)

http://coliru.stacked-crooked.com/a/a4545c398a590b65

[Dkob1]

Ok I see sort of.. but if I want to choose what I want rate and time to be, would I put them equal to 0.0 or something?

Also I want total to equal that part like total = pow(exp(1.0), rate*time) too so what happens to total?
Sorry I'm a beginner, thanks.

[MiiNiPaa]

if I want to choose what I want rate and time to be

You will assign anything you want to them anywhere you want.

[Dkob1]

So I did something like this and it failed.

#include <iostream> 
#include <cmath> 
#include <iomanip> 
using namespace std;

int main()
{
    //declare identifiers 
    double rate = 1; 
    double time = 2;
    float dollars;
    float total=0.0; 

    //initialize identifiers
    total= pow(exp(1.0), rate*time); 

     cout << fixed << showpoint << setprecision(2);
     
    //Enter values
      cout << "Enter the money you want to deposit now(P): " << dollars;
      cout << "Enter the interest rate: " << rate;
      cout << "Enter duration in year(t): " << time;
      cout << "   Your balance after 5 years is " << total; 

	cout << endl; 

	//terminate program
	system("pause");
	return 0;
}

[MiiNiPaa]

It works exactly as written.

//total= pow(exp(1.0), rate*time); 
//total= pow(e, 2*1); 
total = 7.38906

You set total to 7.38906 on line 15 and did not change it after that. So it still equals 7.38906 on line 23 and output will be 7.38906

[Dkob1]

So for what I need to do is choose a number, in my case the rate has to be 0.5 and the time has to be 5 and I have to input that and it asks enter the amount. So that's where I'm stuck.

[MiiNiPaa]

Maybe you want to calculate and write down result after you will know data you need?

int a, b, c;
c = a + b; //Gives you some random data
std::cin >> a >> b;
c = a + b; //Gives you sum of input a abd b 

[Dkob1]

I understand it partially.

[MiiNiPaa]

I noticed, you never actually read data, so this might be helpful: http://www.cplusplus.com/doc/tutorial/basic_io/

[ne555]

> pow(exp(1.0), rate*time);
Shirley you can't be Sirius.

http://www.cplusplus.com/reference/cmath/exp/

Returns the base-e exponential function of x, which is e raised to the power x: e^x.