# Find the ln(X) and log<sub>10</sub>X with the help of expansion

Given a positive number x, the task is to find the natural log (ln) and log to the base 10 (log10) of this number with the help of expansion.

Example:

```Input: x = 5
Output: ln 5.000 = 1.609
log10 5.000 = 0.699

Input: x = 10
Output: ln 10.000 = 2.303
log10 10.000 = 1.000
```

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

Approach:

1. The expansion of natural logarithm of x (ln x) is:

2. Therefore this series can be summed up as:
3. Hence a function can be made to evaluate the nth term of the sequence for 1 &leq; x &leq; n
4. Now to calculate log10 x, below formula can be used:

Below is the implementation of the above approach:

 `// CPP code to Find the ln x and ` `// log10 x with the help of expansion ` ` `  `#include ` `#include ` `#include ` ` `  `using` `namespace` `std; ` ` `  `// Function to calculate ln x using expansion ` `double` `calculateLnx(``double` `n) ` `{ ` ` `  `    ``double` `num, mul, cal, sum = 0; ` `    ``num = (n - 1) / (n + 1); ` ` `  `    ``// terminating value of the loop ` `    ``// can be increased to improve the precision ` `    ``for` `(``int` `i = 1; i <= 1000; i++) { ` `        ``mul = (2 * i) - 1; ` `        ``cal = ``pow``(num, mul); ` `        ``cal = cal / mul; ` `        ``sum = sum + cal; ` `    ``} ` `    ``sum = 2 * sum; ` `    ``return` `sum; ` `} ` ` `  `// Function to calculate log10 x ` `double` `calculateLogx(``double` `lnx) ` `{ ` `    ``return` `(lnx / 2.303); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``double` `lnx, logx, n = 5; ` `    ``lnx = calculateLnx(n); ` `    ``logx = calculateLogx(lnx); ` ` `  `    ``// setprecision(3) is used to display ` `    ``// the output up to 3 decimal places ` ` `  `    ``cout << fixed << setprecision(3) ` `         ``<< ``"ln "` `<< n << ``" = "` `         ``<< lnx << endl; ` `    ``cout << fixed << setprecision(3) ` `         ``<< ``"log10 "` `<< n << ``" = "` `         ``<< logx << endl; ` `} `

 `// Java code to Find the ln x and ` `// log10 x with the help of expansion ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `     `  `// Function to calculate ln x using expansion ` `static` `double` `calculateLnx(``double` `n) ` `{ ` `    ``double` `num, mul, cal, sum = ``0``; ` `    ``num = (n - ``1``) / (n + ``1``); ` ` `  `    ``// terminating value of the loop ` `    ``// can be increased to improve the precision ` `    ``for` `(``int` `i = ``1``; i <= ``1000``; i++)  ` `    ``{ ` `        ``mul = (``2` `* i) - ``1``; ` `        ``cal = Math.pow(num, mul); ` `        ``cal = cal / mul; ` `        ``sum = sum + cal; ` `    ``} ` `    ``sum = ``2` `* sum; ` `    ``return` `sum; ` `} ` ` `  `// Function to calculate log10 x ` `static` `double` `calculateLogx(``double` `lnx) ` `{ ` `    ``return` `(lnx / ``2.303``); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``double` `lnx, logx, n = ``5``; ` `    ``lnx = calculateLnx(n); ` `    ``logx = calculateLogx(lnx); ` `     `  `    ``// setprecision(3) is used to display ` `    ``// the output up to 3 decimal places ` `     `  `    ``System.out.println (``"ln "` `+ n + ``" = "` `+ lnx ); ` `    ``System.out.println (``"log10 "` `+ n + ``" = "``+ logx ); ` `} ` `} ` ` `  `// This code is contributed by ajit  `

 `# Python 3 code to Find the ln x and ` `# log10 x with the help of expansion ` `# Function to calculate ln x using expansion ` `from` `math ``import` `pow` `def` `calculateLnx(n): ` `    ``sum` `=` `0` `    ``num ``=` `(n ``-` `1``) ``/` `(n ``+` `1``) ` ` `  `    ``# terminating value of the loop ` `    ``# can be increased to improve the precision ` `    ``for` `i ``in` `range``(``1``, ``1001``, ``1``): ` `        ``mul ``=` `(``2` `*` `i) ``-` `1` `        ``cal ``=` `pow``(num, mul) ` `        ``cal ``=` `cal ``/` `mul ` `        ``sum` `=` `sum` `+` `cal ` ` `  `    ``sum` `=` `2` `*` `sum` `    ``return` `sum` ` `  `# Function to calculate log10 x ` `def` `calculateLogx(lnx): ` `    ``return` `(lnx ``/` `2.303``) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``n ``=` `5` `    ``lnx ``=` `calculateLnx(n) ` `    ``logx ``=` `calculateLogx(lnx) ` ` `  `    ``# setprecision(3) is used to display ` `    ``# the output up to 3 decimal places ` ` `  `    ``print``(``"ln"``, ``"{0:.3f}"``.``format``(n),  ` `           ``"="``, ``"{0:.3f}"``.``format``(lnx)) ` `    ``print``(``"log10"``, ``"{0:.3f}"``.``format``(n),  ` `              ``"="``, ``"{0:.3f}"``.``format``(logx)) ` `     `  `# This code is contributed by ` `# Surendra_Gangwar `

 `// C# code to Find the ln x and ` `// log10 x with the help of expansion ` `using` `System; ` `     `  `class` `GFG  ` `{ ` `     `  `// Function to calculate ln x using expansion ` `static` `double` `calculateLnx(``double` `n) ` `{ ` `    ``double` `num, mul, cal, sum = 0; ` `    ``num = (n - 1) / (n + 1); ` ` `  `    ``// terminating value of the loop ` `    ``// can be increased to improve the precision ` `    ``for` `(``int` `i = 1; i <= 1000; i++)  ` `    ``{ ` `        ``mul = (2 * i) - 1; ` `        ``cal = Math.Pow(num, mul); ` `        ``cal = cal / mul; ` `        ``sum = sum + cal; ` `    ``} ` `    ``sum = 2 * sum; ` `    ``return` `sum; ` `} ` ` `  `// Function to calculate log10 x ` `static` `double` `calculateLogx(``double` `lnx) ` `{ ` `    ``return` `(lnx / 2.303); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main (String[] args)  ` `{ ` `    ``double` `lnx, logx, n = 5; ` `    ``lnx = calculateLnx(n); ` `    ``logx = calculateLogx(lnx); ` `     `  `    ``// setprecision(3) is used to display ` `    ``// the output up to 3 decimal places ` `     `  `    ``Console.WriteLine(``"ln "` `+ n + ``" = "` `+ lnx ); ` `    ``Console.WriteLine(``"log10 "` `+ n + ``" = "``+ logx ); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

Output:
```ln 5.000 = 1.609
log10 5.000 = 0.699
```

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