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.

Actually, it's $\log_2 6$.

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.

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 $(2^a + 3^b)_{min} = \log_2 3 + \log_2 2 = \log_2 (2\cdot 3) = \log_2 6$

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 $2^{2^{a+b}}=3$

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.

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.