Question 375426
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Hi
Log(4x^2+4x+101)=2
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ log_b(x) \ = \ y \ \ \Rightarrow\ \ b^y = x]
10^2 = 4x^2 + 4x + 101
100 = 4x^2 + 4x + 101
0 = 4x^2 + 4x + 1
factoring
(2x + 1)(2x + 1)= 0 Note: the SUM of inner product(2x) and outer product(2x) = 4x
2x + 1 = 0
x = (-1/2)
there is only one real solution