Lennie Davis

Answered

2022-01-21

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

Answer & Explanation

eskalopit

Expert

2022-01-21Added 31 answers

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

Kayla Kline

Expert

2022-01-22Added 37 answers

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
${\mathrm{log}}_{4}7-\frac{3}{2}>{\mathrm{log}}_{5}4-1.(\ast )$
Now,
$(\ast )\iff 2{\mathrm{log}}_{4}7-3>2{\mathrm{log}}_{5}4-2\iff {\mathrm{log}}_{2}7-3>{\mathrm{log}}_{5}16$
$-2\iff {\mathrm{log}}_{2}7>{\mathrm{log}}_{5}16+1$
$\iff {\mathrm{log}}_{2}7>{\mathrm{log}}_{5}16+{\mathrm{log}}_{5}5\iff {\mathrm{log}}_{2}7>{\mathrm{log}}_{5}80$
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:
${7}^{4}-{2}^{11}=({2}^{3}-1)4-{2}^{11}={2}^{12}-4\cdot {2}^{9}+6\cdot {2}^{6}-4\cdot {2}^{3}+1-{2}^{11}=$
$={2}^{12}-{2}^{11}+3\cdot {2}^{7}-{2}^{5}+1-{2}^{11}=3\cdot {2}^{7}-{2}^{5}+1=3\cdot 128-32+1=353>0.$
Finally, since 80^4=5^4\cdot2^{16} we only need to show that ${5}^{7}>{2}^{16}$ . But
${5}^{7}-{2}^{16}=({5}^{3}{)}^{2}\cdot 5-({2}^{7}{)}^{2}\cdot 4={125}^{2}\cdot 5-{128}^{2}\cdot 4={125}^{2}\cdot 5-(125+3{)}^{2}\cdot 4=$
$={125}^{2}\cdot 5-{125}^{2}\cdot 4-24\cdot 125-36={125}^{2}-24\cdot 125-36=125\cdot (125-24)-36=$
$=125\cdot 101-36=12625-36=12589>0.$

RizerMix

Expert

2022-01-27Added 437 answers

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

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