Question
Solve: 4(log(base2)x)^4 – 7(log(base2)x)^2 +3(log(base2)x) = 0
Answer
Let log(base2)x = y. Then your equation becomes the quartic :
4y^4 – 7y^2 + 3y = 0
y ( 4y^3 – 7y + 3 ) = 0
y ( y -1 ) ( 4y^2 + 4y – 3 ) = 0
y ( y -1 ) ( 2y + 3 ) ( 2y – 1 ) = 0
So,
y = 0 , 1 , (- 3/2), or (1/2)
Hence since x = 2^y,
x = 2^0 , 2^1, 2^(-3/2) or 2^(1/2)
equivalently,
x = 1, 2, 2^(-3/2) or √2.
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