Let `f(x, k) =sum_(n=1)^k log_x(n/x)` , then value of ? which is satisfying the equation `f(x, 2017)= f(x, 2018)` is

Suppose,f(x,n)=sum_(k=1)^(n)log_(x)((k)/(x)), then the value ofx satisfying the equation f(x,10)=f(x,11), is

let f(x)=x^(2)-3x+4 the values of x which satisfies f(1)+f(x)=f(1)*f(x) is

f(x) = log((1+x)/(1-x)) satisfies the equation: f(x_(1)) +f(x_(2))

Let f(x)=(kx)/(x+1) then the value of k such that f(x) is

The value of natural number 'a' for which sum_(k=1)^(n)f(a+k)=16(2^(n)-1) , where the function satisfies the relation f(x+y)=f(x).f(y) for all natural numbers x, y and further f(1)=2 is

If f(x)=sum_(n=0)^(oo)(x^(n))/(n!)(log a)^(n), then at x=0,f(x)

Find the natural number a for which sum_(k=1)^(n)f(a+k)=16(2^(n)-1), where the function f satisfies the relation f(x+y)=f(x)f(y) for all natural number x,y and,further,f(1)=2.

Find the natural number a for which sum_(k=1) ^(n)f(a+k)= 16(2^(n)-1) where the function f satisfies the relation f(x+y)=f(x)f(y) for all natural numbers x,y and further f(1)=2 .