Let $f$ and g be functions satisfying$f(x + y) = f(x)f(y),f(l) = 7$ and $g(x + y) = g(xy)$, $g(l) =

Question

Let $f$ and g be functions satisfying$f(x + y) = f(x)f(y),f(l) = 7$ and $g(x + y) = g(xy)$, $g(l) = 1$, for all $x,y \in \mathbb{N}.\sum_{x = 1}^{n}\mspace{2mu}\left( \frac{f(x)}{g(x)} \right) = 19607$, then n is equal to :

Accepted Answer

$f(x + y) = f(x) \cdot f(y) \Rightarrow f(x) = a^{x}$

$$\left( \because f(1) = 7 \Rightarrow = a^{1} = 7 \right) $$So $f(x) = 7^{x}$

Now

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

$$\Rightarrow g(x + 1) = g(x) $$so $g(1) = g(2) = g(3) = \ldots = g(n) = 1$

Given $\sum_{x = 1}^{n}\mspace{2mu}\frac{f(x)}{g(x)} = 19607$

$${\sum_{x = 1}^{n}\mspace{2mu}\frac{7^{x}}{1} = 19607 }{\Rightarrow 7\left( \frac{7^{n} - 1}{7 - 1} \right) = 19607 }{7^{n} - 1 = \frac{6}{7} \times 19607 }{7^{n} = 16807 \Rightarrow n = 5}$$

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