Let f and g be functions satisfying f(x+y)=f(x) f(y), f(1)=7 and g(x+y)=g(x y), g(1)=1, fo…

Question

Let $f$ and $g$ be functions satisfying $f(x+y)=f(x) f(y), f(1)=7$ and $g(x+y)=g(x y), g(1)=1$, for all $x, y \in \mathbf{N}$. If $\sum\limits_{x=1}^{\mathrm{n}}\left(\frac{f(x)}{\mathrm{g}(x)}\right)=19607$, then n is equal to :

Accepted Answer

$\begin{gathered} f(x+y)=f(x)\cdot f(y)\Rightarrow f(x)={a}^{x} \ (\because f(1)=7\Rightarrow {a}^{1}=7) \end{gathered}$

So $f(x)={7}^{x}$

Now

$g(x+y)=g(xy)(y=1)\Rightarrow g(x+1)=g(x)$

So $g(1)=g(2)=g(3)=\ldots \ldots \ldots =g(n)=1$

$\begin{gathered} \sum _{x=1}^{n}\frac{f(x)}{g(x)}=19607 \ \sum _{x=1}^{n}\frac{{7}^{x}}{1}=19607\Rightarrow 7(\frac{{7}^{n}-1}{7-1})=19607 \ {7}^{n}-1=\frac{6}{7}	imes 19607 \ {7}^{n}=16807\Rightarrow n=5 \end{gathered}$

Solutions · 1