Exponents

Here,

x+x= x^x

Then,

2x= x^x

Or,

x^x= 2x

x^(x-1)= 2^1

i.e. x^(x-1)= 2^(2-1)

Comparing we get,

x=2

$x + x = x^{x} \\ log(x)+log(x)=log(x^{x})\\ 2 \times log(x)=log(x^{x}) \\ 2 \times log(x)=x \times log(x) \\ \frac{2 \times log(x)}{log(x)}=x \\ 2 = x$

If we rewrite that expression in $x + x = 2x$ , we have to satisfy the equation $x^x = 2x$ . By definition, a exponential must have the following form: $x^x = xxxxx...$ , with all terms with the same value. So, with the condition $x > 1$ , the value of $x$ must be equal to 2.

In this question, we have to use some trial and error method tricks. If we use some logic, we can certifying that number must be either 1or 2 because of they are the only options. Then by he equation we have to suggest the right answer.

2 is a very special number

If x is a real numbr , the answer is 2 . But if x is not a real numbr , the answer is 0..

Here,

x+x= x^x

Then,

2x= x^x

Or,

x^x= 2x

x^(x-1)= 2^1

i.e. x^(x-1)= 2^(2-1)

Comparing we get,

x=2. RJ