Given that a>0, b>0, log2^1/2 is the arithmetic mean of log4^a and log2^b. The minimum value of 2/a + 1/b?

Hello, Trefoil2727!

Then I've made only partial progress.

$\displaystyle \text{Given that }a,b>0\text{ and }\log_2\tfrac{1}{2}\text{ is the arithmetic mean of }\log_4a\text{ and }\log_2b$
. . . $\displaystyle \text{find the minimum value of }\tfrac{2}{a} + \tfrac{1}{b}$

We have: .$\displaystyle \frac{\log_4a + \log_2b}{2} \:=\:\log_2\tfrac{1}{2} \quad\Rightarrow\quad \frac{\frac{1}{2}\log_2a + \log_2b}{2} \:=\:-1$

. . . . . . . . $\displaystyle \tfrac{1}{2}\log_2a + \log_2b \:=\:-2 \quad\Rightarrow\quad \log_2a + 2\log_2b \:=\:-4$

. . . . . . . . $\displaystyle \log_2a + \log_2b^2 \:=\:-4 \quad\Rightarrow\quad \log_2(ab^2) \:=\:-4$

. . . . . . . . $\displaystyle ab^2 \:=\:2^{-4} \quad\Rightarrow\quad ab^2 \:=\:\tfrac{1}{16}$

Now what?

emm, is lg 2^(1/2), lg 4^a, lg 2^b

Hello, Trefoil2727!

$\displaystyle \text{Given: }\,a,b>0\text{ and }\log2^{\frac{1}{2}}\text{ is the arithmetic mean of }\log4^a\text{ and }\log2^b.$
$\displaystyle \text{Find the minimum value of: }\,\frac{2}{a} + \frac{1}{b}$

We have: .$\displaystyle \frac{\log4^a + \log2^b}{2} \:=\:\log2^{\frac{1}{2}} \quad\Rightarrow\quad \frac{\log\left(4^a2^b\right)}{2} \:=\:\tfrac{1}{2}\log 2$

. . . . . . . . $\displaystyle \log\left(2^{2a}2^b\right) \;=\;\log 2 \quad\Rightarrow\quad \log\left(2^{2a+b}\right) \;=\;2^1$

. . . . . . . . $\displaystyle 2a+b \:=\:1$

Now what?