Let f and g be functions satisfying f(x+y)=f( | vedclass

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(B) Given $f(x+y)=f(x)f(y)$ with $f(1)=7$. This is a functional equation of the form $f(x)=a^x$. Since $f(1)=7$, we have $a^1=7$, so $f(x)=7^x$. Given

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$5$

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(B) Given $f(x+y)=f(x)f(y)$ with $f(1)=7$. This is a functional equation of the form $f(x)=a^x$. Since $f(1)=7$, we have $a^1=7$, so $f(x)=7^x$. Given $g(x+y)=g(xy)$ with $g(1)=1$. Putting $y=1$, we get $g(x+1)=g(x)$. Since $g(1)=1$, it follows that $g(x)=1$ for all $x \in \mathbb{N}$. The given summation is $\sum_{x=1}^{n} \frac{f(x)}{g(x)} = 19607$. Substituting the functions, we get $\sum_{x=1}^{n} \frac{7^x}{1} = 19607$. This is a geometric progression: $7 + 7^2 + \dots + 7^n = 19607$. The sum of a $GP$ is $S_n = a\frac{r^n-1}{r-1}$. Here $a=7$ and $r=7$. $7 \left(\frac{7^n-1}{7-1}ight) = 19607$. $7 \left(\frac{7^n-1}{6}ight) = 19607$. $7^n-1 = \frac{19607 	imes 6}{7} = 2801 	imes 6 = 16806$. $7^n = 16807$. Since $7^5 = 16807$, we have $n=5$.

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

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