A logical function problem

Question

f(a,n)=nth power of a .If f(a,n-1)=f(a,n)-1,then a+n=? My try: I tried to form an equation and proceed with that further.But I ended up with an absurd conclusion!Here is what I did: Here, f(a,n)=a^n ... ... ... ... ...(i) So, f(a,n-1)=a^(n-1)... ... ...(ii) As the problem says, f(a,n-1)=f(a,n)-1 or, a^(n-1)=(a^n) -1 or, log a based [a^(n-1)]=log a based [(a^n) -1] ;(By taking a based log on both sides of the equation) or, log a based [a^(n-1)]=log a based (a^n) -log a based(1) or, n-1=n-0 or, -1=0 I don't know what is the correct sollution and whether I did any mistake or not .Please help! Thanks in advance.

Answer 1

The question is unclear, but I'll guess that $a,n$ are required to be natural numbers. Absent an assumption of that form there is unlikely to be a unique solution.

First note that we can rule out the case where both $a,n$ are $>1$. Indeed in that case we'd deduce that $a\,|\,1$, a contradiction.

If $a=1$ the equation is impossible.

If $n=1$ we get $1=a^1-1\implies a = 2$.

Thus, $(a,n)=(2,1)$ is the only solution, and we get $a+n=3$.