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2^{2x} \cdot 2^{3\{x\}} = 11 \cdot 2^{5\{x\}} + 5 \cdot 2^{2[x]}

Source: CRMO 2012 region 5 p3 Mumbai

September 30, 2018
algebranumber theoryfractional partsolve- integer partInteger Partsolve- fractional part

Problem Statement

Solve for real xx : 22x⋅23{x}=11⋅25{x}+5⋅22[x]2^{2x} \cdot 2^{3\{x\}} = 11 \cdot 2^{5\{x\}} + 5 \cdot 2^{2[x]}
(For a real number x,[x]x, [x] denotes the greatest integer less than or equal to x. For instance, [2.5]=2[2.5] = 2, [−3.1]=−4[-3.1] = -4, [π]=3[\pi ] = 3. For a real number x,{x}x, \{x\} is defined as x−[x]x - [x].)
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