Number of Digits in a^b

Given two positive integers a and b, task is to find the number of digits in a^b (a raised to the power b).

Example:

Input: a = 2  b = 5
Output: no. of digits = 2
Explanation:
2^5 = 32 
Hence, no. of digits = 2

Input: a = 2  b = 100
Output: no. of digits = 31
Explanation:
2^100 = 1.2676506e+30
Hence, no. of digits = 31

Approach:
The number of digits in a^b can be calculated using the formula:

Number of Digits = 1 + b * (log10a)

When a number is divided by 10, it is reduced by 1 digit.
Example:

554 / 10 = 55, 55 / 10 = 5

Notice, 554 initially has 3 digits but after division there are 2 digits 55 and after further division there is only 1 digit 5. So it can be concluded that to count number of digits, how many times a number is divided by 10 to reach 1 needs to be calculated.
log base 10 of a number is the number of times a number needs to be divided by 10 to reach 1 but as 1 itself is not included in log base 10, 1 is added to get the number of digits.
Note: Floor value of b * (log10a) is taken.

Below is the implementation to calculate the number of digits in a^b.

CPP

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// CPP Program to calculate 
// no. of digits in a^b
#include<iostream>
#include<math.h>
using namespace std;
  
// function to calculate number
// of digits in a^b
int no_of_digit(int a, int b)
{
    return ((int)(b * log10(a)) + 1);
}
      
// driver program
int main()
{
    int a = 2, b = 100;
    cout <<"no. of digits = "<<
                  no_of_digit(a, b);
}
  
// This code is contributed by Smitha

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Java

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// Java Program to calculate 
// no. of digits in a^b
class GFG {
      
    // function to calculate number
    // of digits in a^b
    static int no_of_digit(int a, int b)
    {
        return ((int)(b * Math.log10(a)) + 1);
    }
      
    // driver program
    public static void main(String[] args)
    {
        int a = 2, b = 100;
        System.out.print("no. of digits = " +
                          no_of_digit(a, b));
    }
}

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Python3

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# Python Program to calculate
# no. of digits in a^b
import math
  
# function to calculate number
# of digits in a^b
def no_of_digit(a, b):
    return ((int)(b * math.log10(a)) + 1)
  
# Driver Program
a = 2
b = 100
print("no of digits = ", no_of_digit(a, b))
  
# This code is contributed by Shrikant13

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C#

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// C# Program to calculate 
// no. of digits in a^b
using System;
  
class GFG {
      
    // function to calculate number
    // of digits in a^b
    static int no_of_digit(int a, int b)
    {
        return ((int)(b * Math.Log10(a)) + 1);
    }
      
    // driver program
    public static void Main()
    {
        int a = 2, b = 100;
        Console.Write("no. of digits = " +
                        no_of_digit(a, b));
    }
}
  
// This code is contributed by Smitha.

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PHP

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<?php
// PHP Program to calculate 
// no. of digits in a^b
  
// function to calculate number
// of digits in a^b
function no_of_digit($a, $b)
{
    return ((int)($b * log10($a)) + 1);
}
      
// Driver Code
$a = 2; $b = 100;
echo("no. of digits = " .no_of_digit($a, $b));
  
// This code is contributed by Ajit.
?>

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Output:

no.of digits = 31


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