Solve for real $x$ : $2^{2x} \cdot 2^{3\{x\}} = 11 \cdot 2^{5\{x\}} + 5 \cdot 2^{2[x]}$ (For a real number $x, [x]$ denotes the greatest integer less than or equal to x. For instance, $[2.5] = 2$, $[-3.1] = -4$, $[\pi ] = 3$. For a real number $x, \{x\}$ is defined as $x - [x]$.)
Problem
Source: CRMO 2012 region 5 p3 Mumbai
Tags: algebra, number theory, fractional part, solve- integer part, Integer Part, solve- fractional part
pco
01.10.2018 09:41
parmenides51 wrote: Solve for real $x$ : $2^{2x} \cdot 2^{3\{x\}} = 11 \cdot 2^{5\{x\}} + 5 \cdot 2^{2[x]}$ (For a real number $x, [x]$ denotes the greatest integer less than or equal to x. For instance, $[2.5] = 2$, $[-3.1] = -4$, $[\pi ] = 3$. For a real number $x, \{x\}$ is defined as $x - [x]$.) Infinitely many solutions : $x_n=n+\frac 15\log_2\left(\frac{5\times 4^n}{4^n-11}\right)$ whatever is positive integer $n\ge 2$
Math-wiz
15.08.2019 14:25
The actual question states rational $x$ Maybe the question has a solution now
parmenides51
15.08.2019 14:42
Math-wiz wrote: The actual question states rational $x$ Maybe the question has a solution now so the Resonance solutions here, have solved another problem ? that was my source
Math-wiz
15.08.2019 15:18
parmenides51 wrote: Math-wiz wrote: The actual question states rational $x$ Maybe the question has a solution now so the Resonance solutions here, have solved another problem ? that was my source They have just stayed the problem incorrectly
MatBoy-123
28.11.2020 07:45
No, it's correct and the answer is x=14/5 [X]=2 and {x}=4/5