Maths Olympiad Prep

Library / /13 of 21

Algebra Difficulty 5.3 AIME, harder Prove it Italy

Problem:

For every real number xx, we denote by [x][x] the "integer part of xx", defined as the largest integer x\leq x. Thus for example we have that [3/2]=1[3/2]=1, [π]=3[\pi]=3, [8]=8[8]=8. Determine how many positive real solutions (>0)(>0) the equation 32x=64[x]32^{x}=64^{[x]} has.

Solution

Solution:

The answer is 4. The equation can be written in the form 25x=26[x]2^{5x}=2^{6[x]}, which is equivalent to 5x=6[x]5x=6[x]. Since x<[x]+1x<[x]+1, we have that 6[x]=5x<5[x]+56[x]=5x<5[x]+5, from which [x]<5[x]<5. Since for every x>0x>0 the integer part [x][x] is an integer 0\geq 0, only the possibilities [x]=0,1,2,3,4[x]=0,1,2,3,4 remain. Substituting these values into the relation already found 5x=6[x]5x=6[x], we obtain that the corresponding values of xx are, respectively, 00, 6/56/5, 12/512/5, 18/518/5, 24/524/5. The solution x=0x=0 must however be discarded, since the text required xx positive. One easily verifies instead that the other four values of xx actually solve the proposed equation, which therefore has four positive real solutions.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.