Ajitesh's function

Since $2f(xy) = f(x)^y + f(y)^x$ holds for all real $x,y$ , it must hold for $y = 1$ .

Thus, $2f(x \cdot 1) = f(x) + f(1)^x \leadsto 2f(x) = f(x) + 2^x \leadsto f(x) = 2^x$ .

Clearly, $f(x)^y+f(y)^x = (2^x)^y + (2^y)^x = 2^{xy} + 2^{xy} = 2 \cdot 2^{xy} = 2f(xy)$ .

Therefore, $f(x) = 2^x$ for all real $x$ is indeed the unique solution.

Thus, $\displaystyle \sum_{i = 1}^{100} f(i) = \displaystyle \sum_{i = 1}^{100} 2^i = \dfrac{2^{101} - 2^1}{2-1} = 2^{101} - 2 = 2^a-2 \leadsto a = \boxed{101}$ .

Let $z$ be any real number . Thus $z$ can be written as $( z \times 1 )$ .

Now, putting $(z \times 1)$ in $2f(xy) = [{f(x)}]^y + [{f(y)}]^x$

$\Rightarrow 2f(z \times 1) = [{f(z)}]^1 + [{f(1)}]^z$

$\Rightarrow 2f(z) - [{f(z)}]^1 = [{f(1)}]^z$

$\Rightarrow f(z) = [{f(1)}]^z$

Now, Since $f(1) = 2$

$\Rightarrow f(z)= 2^{z}$

Now,

Let $\ S = \sum_{i=1}^{100} f(i)$

Therefore $\ S = f(1) + f(2) + \dots + f(100)$

$\Rightarrow S = 2^{1} + 2^{2} + \dots + 2^{100}$

$\Rightarrow \frac{S}{2} = 1 + 2^{1} + \dots + 2^{99}$

$\Rightarrow \frac{S}{2} = 1 + S - 2^{100}$

$\Rightarrow 2^{100} = 1 + \frac{S}{2}$

$\Rightarrow 2(2^{100} - 1) = S$

$\Rightarrow 2^{101} - 2 = S$

Hence by comparing $2^{101} - 2$ by $2^a-2$ , we get $a = 101$

Suppose that $y=1$ . We simplify: $2f(x)=f(x)+2^x\implies f(x)=2^x$ . Now we sum $2^1+2^2+\cdots+2^{100}$ . We can use a trick to find the value of this. Let $S=2^1+2^2+\cdots +2^{100}$ . That means $\dfrac{S}{2}=2^0+2^1+\cdots +2^{99}$ . We subtract this from our original equation to get $\dfrac{S}{2}=2^{100}-1$ and therefore $S=2^{101}-2$ so our answer is $\boxed{101}$

For $y = 1$ , we have $2f(x) = f(x) + 2^x \implies f(x) = 2^x$ . Thus the geometric series must be equal to $2\frac{2^{100} - 1}{2-1} = 2^{101} - 2.$ Thus our answer is $101$

Fix $x$ at $1$ . This creates $2f(y) = 2^y + f(y) \Rightarrow f(y) = 2^y$ .

Note that $2^{n+1} - 2= \sum_{i = 1}^n 2^n$ . Thus $\sum_{i = 1}^{100} f(i) = \sum_{i = 1}^{100} 2^i = 2^{101} - 2 \Rightarrow a = 101$ .

Set $X=i$ and $Y=1$

So we find the function as

$f(i)=2^i$

and sum of geometry is $\frac {a(r^n -1)} {r-1}$

for n=100, so $S=2^{101} - 2$

The answer is 101

f (1)=2

Let's y be equal to 1

2f(x)=f(x)+f(1) $^x$

f(x)=2 $^x$

So the sum asked in the question is a geometric progression:

2 $^1$ +2 $^2$ +...+2 $^{100}$

The formula for the sum of geometric progression is:

[2(2 $^{100}$ -1)]/(2-1)

Therefore the result is:

2 $^{101}$ -2

Since that $f(1) = 2$ , we can use the formation law with $x=1$ and $y=n$ . Then, we have:

$2f(n) = f(1)^n + f(n)$ , or, $f(n) =2^n$ .

Therefore, $\sum^{100}_{i=1}f(i) =\sum^{100}_{i=1}2^i = \frac{2(2^{100}-1)}{2-1} = 2^{101}-2$

Thus, $\alpha = 101$

In the given equation put y =1

You get

$2f(x) = f(x) + 2^x$

$f(x) = 2^x$ Hence using summation formula of Geometric progression you get ur required answer as

$2^{101} - 2$

as 2f(xy)=[f(x)]^y+[f(y)]^x holds for all real x,y, it also true for y=1 Then 2f(x.1)=[f(x)]^1+[f(1)]^x 2f(x)=f(x)+2^x [as f(1)=2] =>f(x)=2^x. Clearly,[f(x)]^y+[f(y)]^x=(2^x)^y+(2^y)^x=2^(xy)+2^(xy)=2⋅2^(xy)=2f(xy). Thus, ∑(i=1 to 100)f(i)=∑(i=1 to100)2^i=[2^101−2^1]/[2−1] =2^101−2= 2^a−2=>a=101

Notice that for any $n$ , we see, by the information given by the problem, that $\begin{aligned} 2f(n) &= 2f(n \cdot 1) \\ &= [f(n)]^1 + [f(1)]^n \\ &= f(n) + 2^n \end{aligned}$ And, from simplification, we can now see that, in general, $f(n) = 2^n.$ Neat! Plugging in the values for the summation that we are asked to find, we find that $\sum_{i=1}^{100} f(i) = 2^1 + 2^2 + 2^3 + \dots + 2^{100}$ . The sum we are trying to find is equal to $\begin{aligned} 2^1 + 2^1 + 2^2 + 2^3 + \dots + 2^{100} - 2^{1} &= 2^2 + 2^2 + 2^3 + 2^4 + \dots 2^{100} - 2^1 \\ &= 2^3 + 2^3 + 2^4 + 2^5 + \dots 2^{100} - 2^1 \\ &= \dots \\ &= 2^{101} - 2 \end{aligned}$ because the sum continues to telescope to the final power of $2$ . Thus, $a = \boxed{101}$ .

Since we have $f(1) =2$ we can expressed $2f(i)=2f(1 \times i)=[f(1)]^i + [f(i)]^1=2^i + f(i)$ $\Rightarrow f(i)=2^i$ Thus, $\sum_{i=1}^{100} f(i) = \sum_{i=1}^{100} 2^i=2^1 +2^2 + \cdots +2^{100}$ This is actually just a geometric series of $100$ terms with a common ratio of $2$ . Therefore, $\sum_{i=1}^{100} f(i) = 2^1 +2^2 + \cdots +2^{100} = \frac{2(2^{100}-1)}{2-1}$ which gives $2^{101}-2 \Rightarrow a=101$

Since function is mapped from R to R it is safe to substitute any real value for x without messing up with the domain/range of original function.

Put $x=1$ and $y=y$ in given relation.

$2f(y) = [f(1)]^{y} + [[f(y)]^{1}$

or, $f(y) = 2^{y}$

so, we have sum = $2+ 2^{2} + 2^{3} + 2^{4} + ... 2^{100}$

which is sum of a Geometric Progression .

applying appropriate G.P formula we can easily get $a=101$

If, $2f(xy)=[f(x)]^{y}+[f(y)]^{x}$ , then let, $x=1$ [ We can take the value of $x$ , because, there is no restriction over $x$ , but note that , then the value of $y$ will be dependent on $x$ ]
Then,
$2f(1.y)=[f(1)]^{y}+[f(y)]^{1}$
$⇒f(y)=2^{y}$ [ It is given that, $f(1)=2$ ]
Then,
$\displaystyle \sum_{i=1}^{100} f(i) = \displaystyle \sum_{i=1}^{100} 2^{i}$
$= 2^{1}+2^{2}+2^{3}+2^{4}+........2^{100}$
$= \frac{2(2^{100}-1)}{2-1}$
$= 2^{101}-2$
So, the value of $a$ should be $101$ [ ANSWER ]

Let $i=xy$ and $x=1$ such that $i=y$ . Hence, $\sum_{i=1}^{100} f(i) = \sum_{y=1}^{100} f(y)$ . $$ $$ $$

But, $2f(y) = {f(1)}^y +{f(y)}^x \Rightarrow f(y)={f(1)}^y ={2}^y$ . Therefore $\sum_{y=1}^{100} f(y)=2(2^{100}-1)=2^{101}-2$ , and $a=101$ .

Put $y=1$ i.e

$2f(x)=[f(x)]^1+[f(1)]^x$

$\because f(1)=2$

$\Rightarrow 2f(x)=f(x)+2^x \Rightarrow f(x)=2^x$ Hence,

$\displaystyle \sum_{i=1}^{100} f(i)=2^1+2^2+2^3....2^{100}$

It is a geometric progression which is fairly easy to solve. The result of summation is $2^{101}-2$ and hence, $\fbox{a=101}$ .

Let $y=1$ : $2f(x)=[f(x)]^1+[f(1)]^x=f(x)+2^x\to f(x)=2^x$ . From geometric series, $\sum_{i=1}^{100}2^i=2\frac{2^{100}-1}{2-1}=2^{101}-2$ . $a=\boxed{101}$

Let, $x=1$ and $y=y$ , which leads to $f(x)=2^x$ . Now this form G.P. :)

Using $y=1$ we have $f(x) = 2^{x}$ . So $\sum_{i=1}^{100}f(x) = 2 + 2^{2} + ... + 2^{100}= 2^{101} - 2$

Note that $2f(n)=2f(n\cdot 1)=f(n)+f(1)^n$ , so $f(n)=2^n$ . Thus we have $$\sum {i=1}^{100}f(i)=\sum {i=1}^{100}2^i.$$ Using the identity $$\sum_{i=0}^n 2^i = 2^{n+1}-1,$$ which can be proved inductively, we get $\boxed{a=101}$

Let $y=1$ , so we have: $2f(x\cdot 1)=f(x)+f(1)^x$ $f(x)=f(1)^x=2^x$ Our sum is $2+4+8+\cdots+2^{100}=2^{101}-2$ The answer is $\boxed{101}$ .

First, use x = 1 to find the function f:

[; \mathbf{2f(1\cdot y)=f(1)^{y}+f(y)^{1}} ;]

Organize:

[; \mathbf{f(y)=2^{y}} ;]

At same time:

[; \mathbf{f(x)=2^{x}} ;]

Let S be the sum:

[; \mathbf{S=\sum_{1}^{100}f(i)= 2^{1}+2^{2}+...+2^{100}} ;]

Multiply both side by a factor [; \mathbf{2} ;]:

[; \mathbf{2\cdot S=2^{2} + 2^{3} +...+ 2^{101}} ;]

Then:

[; \mathbf{2S - S = S = 2^{101}-2^{1}} ;]

By the identity:

[; \mathbf{\alpha = 101} ;]