[x]+[2x]+[4x]+[8x]+[16x]+[32x]=12345. What is the real value of x for which the above equation holds true.

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[x] + [2 x] + [4 x] + [8 x] + [16 x] + [32 x] = 12345

x^{3} - 5 x^{2} + 2 x - 10 = 0

x

x_{1}, x_{2}, . . . . ., x_{6}

x^{6} + 2 x^{5} + 4 x^{4} + 8 x^{3} + 16 x^{2} + 32 x + 64 = 0

(1 - y) (1 + 2 x + 4 x^{2} + 8 x^{3} + 16 x^{4} + 32 x^{5}) = (1 - y^{6})

\neq

x

\frac{8 x^{2} + 16 x - 51}{(2 x - 3) (x + 4)} < 3

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