# Multiplication with a power of 2

Given two numbers x and n, we need to multiply x with 2n

Examples :

```Input  : x = 25, n = 3
Output : 200
25 multiplied by 2 raised to power 3
is 200.

Input : x = 70, n = 2
Output : 280
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to compute n-th power of 2 and then multiply with x.

## C++

 `// Simple C/C++ program  ` `// to compute x * (2^n) ` `#include ` `using` `namespace` `std; ` `typedef` `long` `long` `int` `ll; ` ` `  `// Returns 2 raised to power n ` `ll power2(ll n) ` `{ ` `    ``if` `(n == 0) ` `        ``return` `1; ` `         `  `    ``if` `(n == 1) ` `        ``return` `2; ` `         `  `    ``return` `power2(n / 2) * ` `                    ``power2(n / 2); ` `} ` ` `  `ll multiply(ll x, ll n) ` `{ ` `    ``return` `x * power2(n); ` `} ` ` `  `// Driven program  ` `int` `main() ` `{ ` `    ``ll x = 70, n = 2; ` `    ``cout<

## Java

 `// Simple Java program  ` `// to compute x * (2^n) ` `import` `java.util.*; ` ` `  `class` `GFG { ` `     `  `    ``// Returns 2 raised to power n ` `    ``static` `long` `power2(``long` `n) ` `    ``{ ` `        ``if` `(n == ``0``) ` `            ``return` `1``; ` `             `  `        ``if` `(n == ``1``) ` `            ``return` `2``; ` `             `  `        ``return` `power2(n / ``2``)  ` `                          ``* power2(n / ``2``); ` `    ``} ` `      `  `    ``static` `long` `multiply(``long` `x, ``long` `n) ` `    ``{ ` `        ``return` `x * power2(n); ` `    ``} ` `     `  `    ``/* Driver program */` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``long` `x = ``70``, n = ``2``; ` `         `  `        ``System.out.println(multiply(x, n)); ` `    ``} ` `} ` `     `  `// This code is contributed by Arnav Kr. Mandal.     `

## Python3

 `# Simple Python program  ` `# to compute x * (2^n) ` ` `  `# Returns 2 raised to power n ` `def` `power2(n): ` ` `  `    ``if` `(n ``=``=` `0``): ` `        ``return` `1` `    ``if` `(n ``=``=` `1``): ` `        ``return` `2` `    ``return` `power2(n ``/` `2``) ``*` `                  ``power2(n ``/` `2``); ` ` `  ` `  `def` `multiply(x, n): ` `    ``return` `x ``*` `power2(n); ` ` `  ` `  `# Driven program  ` `x ``=` `70` `n ``=` `2` `print``(multiply(x, n)) ` ` `  `# This code is contributed by Smitha Dinesh Semwal `

## C#

 `// Simple C# program  ` `// to compute x * (2^n) ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// Returns 2 raised to power n ` `    ``static` `long` `power2(``long` `n) ` `    ``{ ` `        ``if` `(n == 0) ` `            ``return` `1; ` `             `  `        ``if` `(n == 1) ` `            ``return` `2; ` `             `  `        ``return` `power2(n / 2)  ` `                        ``* power2(n / 2); ` `    ``} ` `     `  `    ``static` `long` `multiply(``long` `x, ``long` `n) ` `    ``{ ` `        ``return` `x * power2(n); ` `    ``} ` `     `  `    ``/* Driver program */` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``long` `x = 70, n = 2; ` `         `  `        ``Console.WriteLine(multiply(x, n)); ` `    ``} ` `} ` `     `  `// This code is contributed by Vt_m.  `

## PHP

 ` `

Output :

`280`

Time complexity : O(Log n)

An efficient solution is to use bitwise leftshift operator. We know 1 << n means 2 raised to power n.

## C++

 `// Efficient C/C++ program to compute x * (2^n) ` `#include ` `typedef` `long` `long` `int` `ll; ` ` `  `ll multiply(ll x, ll n) ` `{ ` `    ``return` `x << n; ` `} ` ` `  `// Driven program to check above function ` `int` `main() ` `{ ` `    ``ll x = 70, n = 2; ` `    ``printf``(``"%lld"``, multiply(x, n)); ` `    ``return` `0; ` `} `

## Java

 `// JAVA Code for Multiplication with a  ` `// power of 2  ` `import` `java.util.*; ` ` `  `class` `GFG { ` `     `  `    ``static` `long` `multiply(``long` `x, ``long` `n) ` `    ``{ ` `        ``return` `x << n; ` `    ``} ` `     `  `    ``/* Driver program to test above function */` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``long` `x = ``70``, n = ``2``; ` `        ``System.out.println(multiply(x, n)); ` `    ``} ` `} ` ` `  `//This code is contributed by Arnav Kr. Mandal. `

## Python3

 `# Efficient Python3 code to compute x * (2^n) ` ` `  `def` `multiply( x , n ): ` `    ``return` `x << n ` `     `  `# Driven code to check above function ` `x ``=` `70` `n ``=` `2` `print``( multiply(x, n)) ` ` `  `# This code is contributed by "Sharad_Bhardwaj". `

## C#

 `// C# Code for Multiplication with a  ` `// power of 2  ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``static` `int` `multiply(``int` `x, ``int` `n) ` `    ``{ ` `        ``return` `x << n; ` `    ``} ` `     `  `    ``/* Driver program to test above function */` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `x = 70, n = 2; ` `     `  `        ``Console.WriteLine(multiply(x, n)); ` `    ``} ` `} ` ` `  `//This code is contributed by vt_m. `

## PHP

 ` `

Output :

`280`

Time complexity : O(1)

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