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

### 2 Answers | Add Yours

**I think the question was supposed to be: x^log 2x +x^log x= 2**

Now in x^log 2x +x^log x= 2 write (x^ log x) as y

=> y^2 + y = 2

=> y^2 + y - 2 = 0

=> y^2 + 2y - y - 2 = 0

=> y( y + 2) - 1(y + 2) = 0

=> (y - 1)(y +2) = 0

So y = 1 and y = -2

x^ log x = 1 and x^ log x = -2

Now x^ log x = -2 is not possible as x would have to be negative and in that case log x would not be defined.

So x^log x = 1

=> log x = 0

=> x = 10^0

=> x = 1

**Therefore x = 1**

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.**