Prove that $\log_27×\log_29<9$

Question

Prove that
$\log_27×\log_29<9$.

I've tried couple of things like multiplying both sides by $\log_28$ Or… moving $\log_29$ to the right which would make everything look like this: $\log_27<\log_9512$

I tried some more things which didn't give anything as well I know this isn't something i should ask (because there isn't anything hard) but my brain doesn't work right now and i don't want to waste that much time on this So i would be thankful if anyone gave somewhat hint or the solution.

Answer 1

Wow, I just loved the problem:

• Using AM-GM inequality

$$\begin{align*} \log_2(7)×\log_2(9) & ≤ \left(\frac{\log_2(7) + \log_2(9)}{2}\right)^2\\ & ≤ \left({\frac{\log_2(63)}{2}}\right)^2\\ & < \left({\frac{\log_2(64)}{2}}\right)^2\\ & < \left({\frac{\log_2({2^6})}{2}}\right)^2\\ & < \left({\frac{3×{2\log_2(2)}}{2}}\right)^2\\ & < 9\left(\frac{\log_2(2)}{1}\right)\\ & < 9 \end{align*}$$

• Using Calculus

$$3 = \log_2(8) = 3×\log_2(2)$$

$$(\log_2{8} = 3) > \log_2(7) = 3-\delta_1$$ $$(\log_2{8} = 3) < \log_2(9) = 3+\delta_2$$ Now here, $\delta_2<\delta_1$

• How $$\frac{d}{dx}\log(x) = \frac{1}{x}$$ $\implies $ for higher values of $x$ the increment in $\log(x)$ starts decreasing.

$$\begin{align*} \log_2(7)×\log_2(9) & = (3-\delta_1)(3+\delta_2)\\ & = 3.3 + 3(\delta_2-\delta_1) -\delta_1\delta_2\\ & = 9+n_1-n_2\\ &= 9-N\\ &<9 \end{align*}$$

Where, $n_1 = 3(\delta_2-\delta_1)$, $n_2 = \delta_2\delta_1$ , and $N>0$

Where $n_1<0$ and $n_2>0$

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Answer 2

Since $(\ln\ln{x})''=-\frac{1+\ln{x}}{x^2\ln^2x}<0$ for $x>1$ and $e^x$ increases, by Jensen we obtain: $$\log_27\log_29=\frac{e^{\ln\ln7+\ln\ln9}}{\ln^22}\leq\frac{e^{2\ln\ln\frac{7+9}{2}}}{\ln^22}=9.$$

Answer 3

As Henry mentioned in the comments:

$$(\log_27)(\log_29) < \left(\frac{\log_27+\log_29}{2} \right)^2 = \left(\frac{\log_2 63}{2} \right)^2 < \left(\frac{\log_2 64}{2} \right)^2 = 3^2=9$$

Alternatively, as $(x-3)(y-3) = xy - 3(x+y) + 9$, we rearrange for $xy$ to get:

$$(\log_27)(\log_29) = (\log_27 - 3)(\log_29 - 3) + 3(\log_27 + \log_29) - 9$$

$$< 0 + 3 \log_2 63 - 9 < 3 \log_2 64 - 9 = 18-9=9.$$

as $\log_27 - 3 < 0, \log_29 - 3 > 0$ between the first and second line.

Answer 4

We have,

$$\begin{align}&\log_27\times\log_29<9\\ \iff&\frac{\log_27}{3}\times \frac{\log_29}{3}<1\\ \iff&\log_87×\log_89<1\\ \iff&\log_98>\log_87.\end{align}$$

It is a well-known fact that, $f(x)=\frac{\ln x}{\ln (x+1)}$ is a strictly increasing function. (see: here)

Thus, you are done.