Question

The equation (x^{​​​​​​x}) raised to x = 2 is satisfied when x is equal to

Answer 1

if your question if x^x=2,Then what is the value of x ?

There is no toher way to solve this other than below mentioned method (which is very difficult to understand at your class std)

This can be solved only if you know logarithms and differentiation which is beyond the scope of your syllabus(You will learn this only in 12 th std)
I will give solution,but you don't understand.

taking log on both sides

log(x^x)=log(2)


=>xlog(x)=log(2) AS log(x^n)=nlog(x)


Differentiating w.r.t to x

=d/dx(xlog(x))=d/dx(log(2)

According to differentiation product rule

=>x*d/dx(log(x)+log(x)*d/dx(x)=0 As d/dx(constant)=0 log(2)-comstant

x*1/x+log(x)*1=0 As d/dx(log(x)=1/x and d/dx(x)=1


1+log(x)=0

=>log(x)=-1


=>x=10^-1 This is also a formula that is if log(a)=b=>a=10^b

x=1/10


x=1/10
=0.1