find x : 10^log^2 x +x^logx= 2

To find x : 10^log^2x +x^logx =
2.

Solution:

 The first term
on LHS: 10^log^2x = 10^(logx)^2.

Second term: Let logx = y.
Then  x = 10^y.

So x^logx = (10^y)^y = 10^y^2 = 10
^(logx)^2.

So LHS = 10^log^2x +x^logx = 10 ^(logx)^2
+10^(logx)^2 = 2*10^(log x)^2.

So the given equation
becomes: 2*10^(logx)^2 = 2.

10^(logx)^2 = 2/2 =
1.

10^(logx)^2 = 1=
10^0.

Since the bases are equal , we equate
exponents:

(logx)^2 = 0.

 logx
= 0.

Therefore x = 1.

Tally:
We put x= 1 in 10^(logx)^2+x^logx = 10^(log1)^2+16log(1) = 10^0^2 + 1^0 = 1+1 = 2=
RHS.

Solution is  x=
1.