Lennie Davis

2022-01-21

Prove ${\mathrm{log}}_{\frac{1}{4}}\frac{8}{7}>{\mathrm{log}}_{\frac{1}{5}}\frac{5}{4}$

eskalopit

Expert

Here is another argument using power-series-based computations of log, but slightly different in its details to Ivans.

Kayla Kline

Expert

Note that ${\mathrm{log}}_{\frac{1}{4}}\frac{8}{7}=-{\mathrm{log}}_{4}\frac{8}{7}={\mathrm{log}}_{4}7-{\mathrm{log}}_{4}8={\mathrm{log}}_{4}7-\frac{3}{2},$ $\mathrm{log}\frac{1}{5}\frac{5}{4}=-{\mathrm{log}}_{5}\frac{5}{4}={\mathrm{log}}_{5}4-{\mathrm{log}}_{5}5={\mathrm{log}}_{5}4-1.$ Hence we need to prove ${\mathrm{log}}_{4}7-\frac{3}{2}>{\mathrm{log}}_{5}4-1.\left(\ast \right)$ Now, $\left(\ast \right)⇔2{\mathrm{log}}_{4}7-3>2{\mathrm{log}}_{5}4-2⇔{\mathrm{log}}_{2}7-3>{\mathrm{log}}_{5}16$ $-2⇔{\mathrm{log}}_{2}7>{\mathrm{log}}_{5}16+1$ $⇔{\mathrm{log}}_{2}7>{\mathrm{log}}_{5}16+{\mathrm{log}}_{5}5⇔{\mathrm{log}}_{2}7>{\mathrm{log}}_{5}80$ I claim that ${7}^{4}>{2}^{11}$ and that ${5}^{11}>{80}^{4}$. Then $4{\mathrm{log}}_{2}7={\mathrm{log}}_{2}{7}^{4}>{\mathrm{log}}_{2}{2}^{11}={\mathrm{log}}_{5}{5}^{11}>{\mathrm{log}}_{5}{80}^{4}=4{\mathrm{log}}_{5}80,$ proving the desired result. Since you explicitly ask not to use a computer let me show the two claimed inequalities by hand: ${7}^{4}-{2}^{11}=\left({2}^{3}-1\right)4-{2}^{11}={2}^{12}-4\cdot {2}^{9}+6\cdot {2}^{6}-4\cdot {2}^{3}+1-{2}^{11}=$ $={2}^{12}-{2}^{11}+3\cdot {2}^{7}-{2}^{5}+1-{2}^{11}=3\cdot {2}^{7}-{2}^{5}+1=3\cdot 128-32+1=353>0.$ Finally, since 80^4=5^4\cdot2^{16} we only need to show that ${5}^{7}>{2}^{16}$. But ${5}^{7}-{2}^{16}=\left({5}^{3}{\right)}^{2}\cdot 5-\left({2}^{7}{\right)}^{2}\cdot 4={125}^{2}\cdot 5-{128}^{2}\cdot 4={125}^{2}\cdot 5-\left(125+3{\right)}^{2}\cdot 4=$ $={125}^{2}\cdot 5-{125}^{2}\cdot 4-24\cdot 125-36={125}^{2}-24\cdot 125-36=125\cdot \left(125-24\right)-36=$ $=125\cdot 101-36=12625-36=12589>0.$

RizerMix

Expert

I will use approximations by power series. It is easy to show that the inequality is equivalent to:$\mathrm{log}\left(3.5\right)\mathrm{log}\left(5\right)>\mathrm{log}\left(4\right)2.$ On the other hand consider the power series expansion for $\mathrm{log}\left(x\right)$ around $3.5$.It is: $\mathrm{log}\left(3.5\right)+\frac{2}{7}\left(x-3.5\right)-\frac{2}{49}\left(x-3.5{\right)}^{2}+O\left(x-3.5{\right)}^{3}$ For $x=5$ we get: $\mathrm{log}\left(5\right)>\frac{33}{98}+\mathrm{log}\left[3.5\right]$ On the other hand for $x=4$ the linear term gives: $\mathrm{log}\left(4\right)<\frac{1}{7}+\mathrm{log}\left(3.5\right)$. Now we aim to prove the following: $\mathrm{log}\left(3.5\right)\mathrm{log}\left(5\right)>\mathrm{log}\left(3.5\right)\cdot \left(\frac{33}{98}+\mathrm{log}\left(3.5\right)\right)>\left(\frac{1}{7}+\mathrm{log}\left(3.5\right){\right)}^{2}>\mathrm{log}\left(4{\right)}^{2}$ The first and last are proved already by the approximations. We need the middle. It is equivalent to the positivity of: $\mathrm{log}\left(3.5\right)\cdot \left(\frac{33}{98}+\mathrm{log}\left(3.5\right)\right)-\left(\frac{1}{7}+\mathrm{log}\left(3.5\right){\right)}^{2}=-\left(\frac{1}{49}\right)+\frac{5}{98}\mathrm{log}\left(3.5\right)=\left(\frac{1}{49}\right)\left(\frac{5}{2}\mathrm{log}\left(3.5\right)-1\right)$ The last is positive because $\mathrm{log}\left(3.5\right)>1>\frac{2}{5}.$