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 $\displaystyle\sum_{x=1}^{n}\left(\dfrac{f(x)}{g(x)}\right) = 19607$, then $n$ is equal to:
A) 6
B) 7
C) 4
D) 5 ✓
✓
Solution Steps
1
Solve for f(n): Cauchy’s Exponential Equation
Given: $f(x+y) = f(x)\cdot f(y)$, $f(1) = 7$
Set $y = 1$: $f(x+1) = f(x)\cdot f(1) = 7\cdot f(x)$
This is a geometric recurrence with ratio 7:
$f(1) = 7,\; f(2) = 49,\; f(3) = 343, \ldots$
$$\boxed{f(n) = 7^n}$$
2
Solve for g(n): Constant Function
Given: $g(x+y) = g(xy)$, $g(1) = 1$
Set $y = 1$: $g(x+1) = g(x\cdot 1) = g(x)$
So $g$ does not change when we add 1. It is constant!
$g(1) = g(2) = g(3) = \cdots = 1$
$$\boxed{g(n) = 1 \text{ for all } n \in \mathbb{N}}$$
3
Simplify the summation
$\dfrac{f(x)}{g(x)} = \dfrac{7^x}{1} = 7^x$
$$\sum_{x=1}^{n} \frac{f(x)}{g(x)} = \sum_{x=1}^{n} 7^x = 7 + 7^2 + \cdots + 7^n$$
This is a geometric series with first term $a=7$, ratio $r=7$, $n$ terms.
4
Apply geometric series formula
$$S = \frac{7(7^n – 1)}{7 – 1} = \frac{7(7^n-1)}{6}$$
Set equal to 19607:
$\dfrac{7(7^n – 1)}{6} = 19607$
$7(7^n – 1) = 117642$
$7^n – 1 = 16806$
$7^n = 16807$
5
Identify n
$7^1 = 7,\quad 7^2 = 49,\quad 7^3 = 343,\quad 7^4 = 2401,\quad 7^5 = 16807$
$7^n = 16807 = 7^5 \implies \boxed{n = 5}$
✓ Verify: $7+49+343+2401+16807 = 19607$ ✓
| x | $7^x$ | Running sum |
|---|---|---|
| 1 | 7 | 7 |
| 2 | 49 | 56 |
| 3 | 343 | 399 |
| 4 | 2401 | 2800 |
| 5 | 16807 | 19607 ✓ |
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Key Insight
f(x+y)=f(x)f(y) → f(n)=7ⁿ. g(x+y)=g(xy) with y=1 → g constant=1. Sum of 7ˣ geometric series = 19607 → n=5.
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Theory
1. Cauchy’s Exponential Functional Equation
$f(x+y) = f(x)f(y)$ is Cauchy’s exponential equation. On natural numbers, with $f(1) = a$: setting $y=1$ gives $f(x+1) = a\cdot f(x)$. This is a geometric sequence, so $f(n) = a^n$. Here $a = 7$, giving $f(n) = 7^n$. Note: $f(x+y) = f(x)+f(y)$ (additive) gives linear functions, but $f(x+y)=f(x)f(y)$ (multiplicative) gives exponential functions.
2. The Constant Function from g(x+y)=g(xy)
Setting $y=1$: $g(x+1) = g(x\cdot 1) = g(x)$. This means $g$ has the same value at consecutive natural numbers — hence $g$ is constant on $\mathbb{N}$. Starting from $g(1)=1$, we get $g(n)=1$ for all $n \in \mathbb{N}$. This clever substitution $y=1$ is the key trick to determine $g$ without solving the full functional equation.
3. Geometric Series Formula
Sum of geometric series: $a + ar + ar^2 + \cdots + ar^{n-1} = \dfrac{a(r^n-1)}{r-1}$ for $r \neq 1$. Here $\sum_{x=1}^n 7^x = 7+7^2+\cdots+7^n = \dfrac{7(7^n-1)}{6}$. Setting this equal to 19607 and simplifying gives $7^n = 16807 = 7^5$, so $n=5$.
4. Powers of 7 — Must Know
$7^1=7$, $7^2=49$, $7^3=343$, $7^4=2401$, $7^5=16807$, $7^6=117649$. In JEE problems involving $f(1)=7$ and geometric sums, recognizing that $16807 = 7^5$ instantly gives $n=5$. Memorizing small powers of 2,3,5,7 saves crucial time in exams.
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FAQs
Q
Why does f(x+y)=f(x)f(y) give f(n)=7^n?
⌄
Set y=1: f(x+1)=f(x)·f(1)=7f(x). This is geometric with ratio 7. Since f(1)=7, f(n)=7^n for all n∈N.
Q
Why does g(x+y)=g(xy) give g(n)=1?
⌄
Set y=1: g(x+1)=g(x·1)=g(x). So g is constant on N. Since g(1)=1, all values are 1.
Q
What is the sum Σ7^x from x=1 to n?
⌄
Geometric series: 7+49+343+…+7^n = 7(7^n-1)/(7-1) = 7(7^n-1)/6.
Q
How to solve 7(7^n-1)/6 = 19607?
⌄
Multiply both sides by 6: 7(7^n-1)=117642. Divide by 7: 7^n-1=16806. Add 1: 7^n=16807=7^5. So n=5.
Q
What is 7^5?
⌄
7^1=7, 7^2=49, 7^3=343, 7^4=2401, 7^5=16807. Memorize these for JEE speed!
Q
Verify that 7+49+343+2401+16807=19607?
⌄
7+49=56. 56+343=399. 399+2401=2800. 2800+16807=19607. ✓ Confirmed.
Q
What is Cauchy’s functional equation?
⌄
f(x+y)=f(x)+f(y) is additive Cauchy (gives f(n)=cn). f(x+y)=f(x)f(y) is exponential Cauchy (gives f(n)=a^n). Here the exponential version applies.
Q
Could g(x+y)=g(xy) have other solutions?
⌄
On N with g(1)=1: setting y=1 forces g to be constant=1. No other solution exists under this constraint for natural numbers.
Q
Why not n=6 or n=7?
⌄
n=6 gives sum=7+…+7^6=7(7^6-1)/6=7(117648)/6=137256. n=5 gives exactly 19607. So n=5 is unique.
Q
What if f(1) were different, say f(1)=3?
⌄
Then f(n)=3^n and the sum becomes Σ3^x = 3(3^n-1)/2. Setting equal to the given value would give a different n. The method is identical.
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