Find the value of real parameter $ a$ such that $ 2$ is the smallest integer solution of \[ \frac{x+\log_2 (2^x-3a)}{1+\log_2 a} >2.\]
Problem
Source: Indonesia TST 2009 Second Stage Test 2 P2
Tags: parameterization, logarithms, inequalities, algebra proposed, algebra
Differ
06.04.2009 00:20
$ x + \log_2(2^x - 3a) > 2(1 + \log_2a)$ $ x + \log_2(2^x - 3a) > \log_2(4a^2)$ $ 2^{2x} - 3a2^x > 4a^2$ $ 2^{2x} - 3a2^x - 4a^2 > 0$ $ (2^x - 4a)(2^x + a) > 0$ We need all $ 2^x \leq 2$ not to work. That gives us $ a \leq -2$, $ a \geq \frac{1}{2}$.
stephencheng
07.04.2009 12:04
Differ wrote: $ x + \log_2(2^x - 3a) > 2(1 + \log_2a)$ $ x + \log_2(2^x - 3a) > \log_2(4a^2)$ $ 2^{2x} - 3a2^x > 4a^2$ $ 2^{2x} - 3a2^x - 4a^2 > 0$ $ (2^x - 4a)(2^x + a) > 0$ We need all $ 2^x \leq 2$ not to work. That gives us $ a \leq - 2$, $ a \geq \frac {1}{2}$. Your method works but you made certain mistakes in your solution. Mistake 1: $ 1+log_2 a$ can be negative, so the first row is wrong. Mistake 2: $ a$ cannot be negative or $ log_2 a$ undefined. So $ a \le -2$ is impossible.
Xantos C. Guin
07.04.2009 18:11
First of all, note that $ x = \frac{1}{2}$ or $ a \le 0$ yield an undefined expression.
Case I: $ 1 + \log_2a > 0 \Rightarrow a > \frac{1}{2}$
$ x + \log_2(2^x - 3a) > 2(1 + \log_2a)$
$ x + \log_2(2^x - 3a) > \log_2(4a^2)$
$ 2^{2x} - 3a2^x > 4a^2$
$ 2^{2x} - 3a2^x - 4a^2 > 0$
$ (2^x - 4a)(2^x + a) > 0$
Both factors are positive if $ 2^x > 4a$ and $ 2^x > -a$, or equivalently $ 2^x > 4a$.
Since $ x \le 1$ does not fit this condition, and $ x=2$ does, we have $ 2^1 \le 4a < 2^2 \Rightarrow a \in \left(\frac{1}{2}, 1\right)$.
Both factors are negative if $ 2^x < 4a$ and $ 2^x < -a$, or equivalently $ 2^x < -a$.
However, this is impossible since $ a > \frac{1}{2} \Rightarrow 2^x < -\frac{1}{2}$.
Case II: $ 1 + \log_2a < 0 \Rightarrow 0 < a < \frac{1}{2}$
$ x + \log_2(2^x - 3a) < 2(1 + \log_2a)$
$ x + \log_2(2^x - 3a) < \log_2(4a^2)$
$ 2^{2x} - 3a2^x < 4a^2$
$ 2^{2x} - 3a2^x - 4a^2 < 0$
$ (2^x - 4a)(2^x + a) < 0$
Since $ 2^x+a > 2^x - 4a$, we must have $ 2^x+a > 0$ and $ 2^x - 4a < 0$.
This is equivalent to $ -a < 2^x < 4a$. However, since $ a>0$ and $ 2^x > 0$, if $ x = 2$ satisfies this inequality, then any $ x \le 1$ also satisfies this inequality.
Thus, there are no permissible values of $ a$ in this case.
Therefore, the only values of $ a$ such that the smallest integer solution is $ x = 2$ is $ a \in \left( \frac{1}{2}, 1 \right)$.
Differ
07.04.2009 22:19
stephencheng wrote: Differ wrote: $ x + \log_2(2^x - 3a) > 2(1 + \log_2a)$ $ x + \log_2(2^x - 3a) > \log_2(4a^2)$ $ 2^{2x} - 3a2^x > 4a^2$ $ 2^{2x} - 3a2^x - 4a^2 > 0$ $ (2^x - 4a)(2^x + a) > 0$ We need all $ 2^x \leq 2$ not to work. That gives us $ a \leq - 2$, $ a \geq \frac {1}{2}$. Your method works but you made certain mistakes in your solution. Mistake 1: $ 1 + log_2 a$ can be negative, so the first row is wrong. Mistake 2: $ a$ cannot be negative or $ log_2 a$ undefined. So $ a \le - 2$ is impossible. And that's why I got bronze on USAMTS.