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