Let $a$ and $b$ be real numbers such that $a+b=\log_2( \log_2 3)$. What is the minimum value of $2^a + 3^b$ ?
Problem
Source: Kosovo MO 2020 Grade 11, Problem 3
Tags: algebra, minimum value, logarithms, national olympiad, Olympiad, Kosovo
02.02.2020 18:08
Circumcircle wrote: Let $a$ and $b$ be real numbers such that $a+b=\log_2( \log_2 3)$. What is the minimum value of $2^a + 3^b$ ? I got the minimum value as $\log_2 6$ Will post solution in a while.
02.02.2020 19:18
Actually, it's $\log_2 6$.
02.02.2020 19:23
TLP.39 wrote: Actually, it's $\log_2 6$.
Exactly it's $\log_2 6$ I didn't resolved my last steps much, so it ended up like that.
02.02.2020 19:34
So it's an instantaneous solution which actually suits to the competition environment with time constraints. Do let me know if you have more short methods to reach the solution. $\log_2(\log_2 3) = a+b$ $2^{a+b} = \log_2 3$ $2^{2^{a+b}}=3$ $(2^{2^{a+b}})^b=3^b$ Now we have $2^a = \frac{\log_2 3}{2^b}$ And $3^b = (2^{2^{a+b}})^b$ $(2^a + 3^b) = \frac{\log_2 3}{2^b} + (2^{2^{a+b}})^b$ So $2^a + 3^b$ is minimum when we have denominator reduced to it's best possible Now plugging $b = 0$ We have $(2^a + 3^b)_{min} = \frac{\log_2 3}{1} + 1$ Which we can write it as $(2^a + 3^b)_{min} = \frac{\log_2 3}{1} + \log_2 2$ Even further it reduces to $(2^a + 3^b)_{min} = \log_2 3 + \log_2 2 = \log_2 (2\cdot 3) = \log_2 6$
03.02.2020 11:34
Alphatrion wrote: $\log_2(\log_2 3) = a+b$ $2^{a+b} = \log_2 3$ $4^{a+b} = 3$ This is wrong. It should be $2^{2^{a+b}}=3$
03.02.2020 12:55
Supercali wrote: Alphatrion wrote: $\log_2(\log_2 3) = a+b$ $2^{a+b} = \log_2 3$ $4^{a+b} = 3$ This is wrong. It should be $2^{2^{a+b}}=3$ Thanks a lot for pointing it out. I'll come up with better solution soon.
03.02.2020 21:26
TLP.39's Solution \begin{align*} 2^a+3^b&=2^{\log_2 (\log_2 3)}2^{-b}+3^b\\ &= (\log_2 3)2^{-b}+3^b\\ &\ge (\log_2 3+1)\left(2^{-\frac{\log_2 3}{\log_2 3+1}}3^{\frac{1}{\log_2 3+1}}\right)^b\\ &=log_2 6 \end{align*} Please elaborate what happened in between the 2nd and 3rd step, how does that transformation took place. How you established the inequality. Rest everything is clear after when we arrive at the 3rd step. So I'll break that into few segments $\log_2 3 + 1 = \log_2 6$ Now $2^{-(\frac{\log_2 3}{\log_2 6})} = 2^{\log_3 6}$ Similarly $3^{(\frac{1}{\log_2 6})} = 3^{\log_6 2}$ Now finally we have to deal with $(2^a + 3^b)_{min} = \log_2 6 \cdot \bigg(2^{\log_3 6} \cdot 3^{\log_6 2}\bigg)^b$ So in conclusion minimum value happens when $b=0$, also it brings no harm to our system, one may go through few of the initial relationships.