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 * (log_{10}a)

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 * (log _{10}a)** is taken.

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

## CPP

`// 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

` ` `// 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

`# 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#

`// 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

`<?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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