Prove \log_{\frac{1}{4}}\frac{8}{7}>\log_{\frac{1}{5}}\frac{5}{4}

Lennie Davis

Lennie Davis

Answered question

2022-01-21

Prove log1487>log1554

Answer & Explanation

eskalopit

eskalopit

Beginner2022-01-21Added 31 answers

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

Kayla Kline

Beginner2022-01-22Added 37 answers

Note that log1487=log487=log47log48=log4732, log1554=log554=log54log55=log541. Hence we need to prove log4732>log541.() Now, ()2log473>2log542log273>log516 2log27>log516+1 log27>log516+log55log27>log580 I claim that 74>211 and that 511>804. Then 4log27=log274>log2211=log5511>log5804=4log580, proving the desired result. Since you explicitly ask not to use a computer let me show the two claimed inequalities by hand: 74211=(231)4211=212429+626423+1211= =212211+32725+1211=32725+1=312832+1=353>0. Finally, since 80^4=5^4\cdot2^{16} we only need to show that 57>216. But 57216=(53)25(27)24=1252512824=12525(125+3)24= =12525125242412536=12522412536=125(12524)36= =12510136=1262536=12589>0.
RizerMix

RizerMix

Expert2022-01-27Added 656 answers

I will use approximations by power series. It is easy to show that the inequality is equivalent to:log(3.5)log(5)>log(4)2. On the other hand consider the power series expansion for log(x) around 3.5.It is: log(3.5)+27(x3.5)249(x3.5)2+O(x3.5)3 For x=5 we get: log(5)>3398+log[3.5] On the other hand for x=4 the linear term gives: log(4)<17+log(3.5). Now we aim to prove the following: log(3.5)log(5)>log(3.5)(3398+log(3.5))>(17+log(3.5))2>log(4)2 The first and last are proved already by the approximations. We need the middle. It is equivalent to the positivity of: log(3.5)(3398+log(3.5))(17+log(3.5))2=(149)+598log(3.5)=(149)(52log(3.5)1) The last is positive because log(3.5)>1>25.

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