João's functional equation

We will attempt to arrive at $f(5)$ through repeated plugging in of $f(x+y)=f(x)f(y)$ .

Since $f(2)=f(1+1)$ we have $f(2)=f(1+1)=f(1)f(1)=9$ ; therefore $f(1)=3$ .

(We cannot have $f(1)=-3$ because the function maps positive integers to positive integers.)

Now we multiply $f(1)f(2)=f(1+2)=f(3)=3\times 9=27$ .

Similarly, multiply $f(2)f(3)=f(2+3)=f(5)=9\times 27=\boxed{243}$ and we are done.

Firstly, it is given- $f(2) = 9$ We can simplify it to get-

$f(1 + 1) = 9$

$f(1)^2 = 9$

$f(1) = 3$ (Since $f(1)$ can't be negative.)

Using it furthur in-

$f(5) = f(4 + 1)$

$f(5) = f(4)f(1)$

$f(5) = f(2 + 2)f(1)$

$f(5) = f(2)f(2)f(1)$

$f(5) = 9 \times 9 \times 3$

Therefore, $f(5) = \fbox{243}$ .

f (x + y) = f(x)f(y) is exposional function, so

f(x + y) = a^(x+y) such that

f(2) = 9

f(2) = 3^2

f(5) = 3^5

f(5) = 243

Starting with $f(2)=9$ , it can be expressed as $f(x+y)=f(x)f(y)$ . Since $x$ and $y$ must be positive, the only two numbers satisfying this are $1$ and $1$ . Plugging in, $f(2)=f(1)f(1)=9$ . Taking the square root, $f(1)=\pm3$ . Since the function is from the positive integers to the positive integers, the positive solution of $3$ is the only acceptable one.

Now for $f(5)$ , it can be expressed as $f(4)f(1)$ . That can be expressed as $f(3)f(1)$ and so on, until $f(2)=f(1)f(1)$ . Substituting in the values, we get $f(5)=[f(1)]^5$ . Substituting for $f(1)$ , $f(5)=3^5$ or $243$ .

And Finally by Induction, we can generalize that f(x)=3^x

f(1+1)=f(1)*f(1)=(f(1))^2

  f(2)=9
  (f(1))^2=9
   f(1)=3
   f(2)=9
   f(3)=f(2+1)=f(2)*f(1)
   =(9)(3)=27

. . . . f(5)=f(4+1) =f(4)*f(1) =f(2)f(2)f(1) =(9) (9)( 3) =243

Since $f(2)=9$ , $f(2)=f(1+1)=f(1)^2=9$ Then, $f(1)=3$ . Now, $f(x+1)=3f(x)$ Then, $f(5)=3f(4)=9f(3)=27f(2)=\boxed{243}$

this condition can only be fulfilled when f(x) is an exponential function ............. that is f(x) = (a)^x .... it is given that f(2) = 9............ therefore (a)^2=9 implies a=3. now calculate ( 3)^5

An easy way to solve this is to start with what you know and work backwards. We know that since 2=1+1 we can substitute to find that $f(2)=f(1)^2$ . Meaning that $f(1)=3$ . Note that f as a function is positive so the square root of 9 must be positive. So, we are going to work backwards. $f(5)=f(3)f(2)$ which is the same as $f(5)=9*f(3)$ . Since $f(3)=f(1)f(2)=27$ we can substitute to find that $f(5)=3^5=243$ .

If f(2) = 9, we may accept that x+y = 2 (property) and as x and y are positive integers, they can on only have the value of one x = y = 1. Then: f(2) = f(1+1) = f(1) f(1) = 9, f(1) f(1) = 9 f(1)^2 = 9 f(1) = 3 (It cannot be -3 because the function is from positive integers to positive integers).

Now, we may decompose f(5) in the next way: f(5) = f(2+2+1) = f(2) f(2) f(1) We know: f(2) = 9 and f(1) = 3 So, f(5) = 9 9 3 f(5) = 243

$f(2)=f(1+1)=f(1)f(1)$ . Thus $f(1)f(1)=f^2(1)=9$ and $f(1)=\sqrt 9=3$ . $f(5)=f(2+2+1)=f(2)f(2)f(1)=9 \times 9 \times 3=243$

You are given that f(2) = 9 This can be rewritten as f(1+1)=9 -> f(1)*f(1)=9 -> f(1)^2=9 -> f(1)=3

You are trying to find f(5) This can be rewritten as f(1)^5

We already found that f(1)=3, so f(1)^5 = 243

f(4) = f(2)f(2) = 9 x 9 = 81

f(4) = f(3)f(1) = f(2)f(1)f(1)

f(1)f(1) = 9

f(1) = 3

f(3) = f(1)f(2) = 3 x 9 = 27

f(5) = f(2)f(3) = 9 x 27 = 243

$f(2) = f(1+1) = 9 = 3 * 3$ . Thus, $f(1) = 3$ and $f(5) = f(1+1+1+1+1) = 3^5 = 243$ .