// Created by Robert Ronge //By Madhav /* [Challenge] Logarithm calculator Design a logarithmic calculator.It must accept 2 inputs the base and the number.. Example: log(2,8) will return 3 log(1.5,2.25)will return 2 log(10,2) will return 0.3013... If no base is provided it must default to mathematical contant e. Bonus:Find a way to do this without using inbuilt log function ,if any. */ #include using namespace std; long double root(long long n, long double x); // definition under main // Approximation of // the natural logarithm: long double ln(long double a) { long long n = 10000000000000; return n * (root(n, a) - 1); }; // The proof: // 1/n // lim( n * ( a - 1 ) ) // n->∞ // Let x be equal to 1/n: // x // a -1 // = lim( ----- ) // x->0 x // Use L’Hôpital’s rule: // d x // --(a -1) // dx // = lim( ---------- ) // x->0 d // --( x ) // dx // Derive both sides: // x // ln(a) * a // = lim( ---------- ) // x->0 1 // Set x to 0: // = ln(a) float e = 0; float log(float x, float &base = e) { if(base == 0) { int n = 1; float f = 1; float temp = -1; while(temp != base) { temp = base; base += f; f /= n; n++; }; }; return ln(x) / ln(base); }; int main() { float base = 0, x; cin >> x >> base; float res = log(x, base); cout << "log" << base << "(" << x << ")=" << res; return 0; }; // x^n: long double pow( long double x, long long n ) { long double res = 1; long long ex = n; if(n < 0) ex = n * -1; while(ex) { if(ex & 1) { res *= x; ex--; }; ex = ex >> 1; x *= x; }; if(n < 0) return 1 / res; return res; }; // n-th root of x: long double root( long long n, long double x ) { long double res = 1; for( long long i = 0; i < 1000000; i++ ) { long double temp = ( x / pow(res, n - 1) - res) / n; if(temp == 0) break; res += temp; }; return res; };

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