Which of the following is bigger (logarithms)

Question

I need to compare those two expressions and decide which is bigger.

$2 \sqrt2$ or $\log_2(3)+\log_3(4) $.

So I tried to simplify so the log expression so I know

and so

$$ \log_2(4) \times (\log_4(3) + \log_3(2)) ?? 2 \times \sqrt2$$

and then

$$2 \times \log_2(2)\times(\log_4(3)+\log_3(2)) ?? 2 \sqrt2$$

$$\log_2(2) \times (\log_4(3)+\log_3(2)) ?? \sqrt2 $$

and I know $\ log_2(2) = 1$ so now I need to compare those two expressions:

$$\log_4(3)+\log_3(2) $$against$$ \sqrt2 $$

I'm not really sure what i'm doing wrong here.

Answer 1

$\log_3 4 = \dfrac{\log_2 4}{\log_2 3} = \dfrac{2}{\log_2 3}$

$A := {\log_2 3}+ \log_3 4 = {\log_2 3} + \dfrac{2}{\log_2 3}$

Dividing by$A$ by $\sqrt 2$, observe $ \dfrac{\log_2 3}{\sqrt 2} + \dfrac{\sqrt 2}{\log_2 3} > 2$ by AM-GM inequality (since ${\log_2 3 \ne \sqrt 2}$)

Thus $A>2\sqrt 2$

Answer 2

$$\log_23+\log_34>2\sqrt2$$ because it's $$\log_23+\frac{2}{\log_23}>2\sqrt2,$$ which is AM-GM: $\log_23+\frac{2}{\log_23}>2\sqrt{\log_23\cdot\frac{2}{\log_23}}=2\sqrt2$ or $$\log^2_23-2\sqrt2\log_23+2>0$$ 0r $$\left(\log_23-\sqrt2\right)^2>0,$$ which is obvious.

Answer 3

We have \begin{eqnarray*} ((\ln 3)^2 -2 (\ln 2)^2)^2 \geq 0 \\ ((\ln 3)^4 - 4 ((\ln 3)^2(\ln 2)^2 + 4(\ln 2)^4 \geq 0 \\ ((\ln 3)^4 + 4 ((\ln 3)^2(\ln 2)^2 + 4(\ln 2)^4 \geq 8 ((\ln 3)^2(\ln 2)^2 \\ \end{eqnarray*} Now sqauare root this ... \begin{eqnarray*} ((\ln 3)^2 +2 (\ln 2)^2 \geq \sqrt{8} (\ln 3)(\ln 2) \\ \frac{ \ln 3}{\ln 2}+ \frac{ 2 \ln 2}{ \ln 3} \geq \sqrt{8} \\ \log_2 3 + \log_3 4 \geq 2 \sqrt{2}. \end{eqnarray*}