Romanian Mathematical Olympiad

Question

a) x R _2[x] = [_2 x] = _m N [2^m, 2^m + 1). b) x R 2^ x = 2^x = _m N [m, _2 (2^m + 1)). (here, [a] denotes the integer part (floor function) of the real number a).

Accepted Answer

a) The existence of the logarithms requires

x > 1

. If

[lo g_{2} x] = m

, then

m \in N

and

2^{m} \leq x < 2^{m + 1}

. From

lo g_{2} [x] = m

, follows

[x] = 2^{m}

, that is

2^{m} \leq x < 2^{m} + 1

. Conversely,

x \in [2^{m}, 2^{m} + 1)

,

m \in N

yields

[lo g_{2} x] = lo g_{2} [x] = m

.

b) If

[2^{x}] = t \in N

, then

lo g_{2} t \leq x < lo g_{2} (t + 1)

. From

2^{[x]} = t

follows

[x] = lo g_{2} t = m \in N

, whence

m \leq x < lo g_{2} (2^{m} + 1)

. Conversely, if

x \in [m, lo g_{2} (2^{m} + 1))

,

m \in N

then

[2^{x}] = 2^{[x]} = m

.