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

How to prove without a computer?

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up vote 4 down vote accepted

Here is another argument using power-series-based computations of $\log$, but slightly different in its details to Ivan's.

$$\log_{1/4} 8/7 = -\log_4 8/7 = \log_4 7/8 = \log_4 (1 - 1/8),$$ while $$\log_{1/5} 5/4 = -\log_5 5/4 = \log_5 4/5 = \log_5 (1 - 1/5).$$

So the problem reduces to showing that $$\log_4(1-1/8) >\log_5 (1- 1/5).$$

This is slightly delicate, because both numbers of which we are taking the $\log$ are close to $1$, so both sides are close to $0$, and we have to estimate how close.

Recall that $$\ln (1- x) = - x - x^2/2 - x^3/3 - \cdots$$ when $|x| < 1$, and that $\log_a (1-x) = (1/\ln a)\ln (1-x)$. So we have to prove $$\ln 4 \, (1/5 + 1/50 + \cdots ) > \ln 5 \,(1/8 + 1/128 + \cdots).$$ Now clearly $$(1/5 + 1/50 + \cdots ) > 8/5 (1/8 + 1/128 + \cdots),$$ so it suffices to show that $$(8/5)\ln 4 > \ln 5,$$ i.e. that $$\ln 4^8 > \ln 5^5,$$ i.e. that $$65536 > 3125,$$ which is true.

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Note that

$\log_{\frac{1}{4}}\frac{8}{7}=-\log_4\frac{8}{7}=\log_47-\log_48=\log_4 7-\frac{3}{2}$,


Hence we need to prove $$\log_47-\frac{3}{2}>\log_54-1.\ (*)$$ Now, $$(*)\Leftrightarrow2\log_47-3>2\log_54-2\Leftrightarrow \log_27-3>\log_516-2\Leftrightarrow\log_27>\log_516+1\Leftrightarrow$$ $$\Leftrightarrow\log_27>\log_516+\log_55\Leftrightarrow\log_27>\log_580.$$

I claim that $7^4>2^{11}$ and that $5^{11}>80^4$. Then $$4\log_27=\log_27^4>\log_22^{11}=\log_55^{11}>\log_580^4=4\log_580,$$ 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 $80^4=5^4\cdot 2^{16}$ we only need to show that $5^7>2^{16}$. But $$5^7-2^{16}=(5^3)^2\cdot 5-(2^7)^2\cdot 4=125^2\cdot 5-128^2\cdot 4=125^2\cdot 5-(125+3)^2\cdot 4=$$ $$=125^2\cdot 5-125^2\cdot 4-24\cdot 125-36=125^2-24\cdot125-36=125\cdot(125-24)-36=$$ $$=125\cdot 101-36=12625-36=12589>0.$$

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My second attempt, hopefully not flawed this time. (Although I think that there must be a simpler way to do this.)

I will use $\ln$ to denote natural logarithm. (Some people use $\log$)

The inequality $$\log_{\frac{1}{4}} \frac{8}{7}> \log_{\frac{1}{5}} \frac{5}{4}$$ can be rewritten as $$\frac{\ln\frac87}{\ln\frac14} > \frac{\ln\frac54}{\ln\frac15}.$$ This is equivalent to the inequality $$\ln\frac87 \ln \frac15 > \ln \frac54 \ln\frac14$$ (we have multiplied both sides of the inequality by $\ln\frac14$ and $\ln\frac15$. So we have multiplied the inequality by a negative number twice - as a final result the inequality signs remains unchanged.)

This leads to equivalent inequalities

$$-\ln 5 \ln\frac87 > -\ln 4 \ln \frac54\\ \ln 4 \ln \frac54 > \ln 5 \ln\frac87. \tag{*}$$

So the above inequality $(*)$ is what we want show.

We know that $\ln(1+x)<x$ and $\ln(1+x)>\frac{x}{1+x}$ for $x>0$. (Both these inequalities can be verified simply by taking derivative of functions $\ln(1+x)-x$ and $\ln(1+x)-\frac{x}{1+x}$ and showing that the derivative is negative/positive for $x>0$. See WolframAlpha: 1, 2. Of course, there can be other ways how to prove this. Both these properties are mentioned at Wikipedia.)

Now if we use one of these inequalities for $x=\frac17$, we get $\ln\frac87 < \frac17$, which gives $$\ln\frac87\ln 5 < \frac17\ln5.$$

For $x=\frac14$ we have $\ln\frac54 > \frac{1/4}{1+1/4} = \frac{1/4}{5/4} = \frac15$ and thus $$\ln \frac54 \ln 4 > \frac15 \ln 4.$$

Now we may notice that the following inequalities are equivalent to each other: $$\frac15 \ln 4 > \frac17\ln5 \tag{**}\\ 7\ln 4 > 5\ln 5\\ \ln(4^7) > \ln{5^5}\\ 4^7 > 5^5\\ $$ The last inequality is true, which can be seen by evaluating $4^7=16384$ and $5^5=3125$. (And this can be shown by hand, too, since $4^7=2^{14}=16\cdot2^{10}=16\cdot1024 > 16000$ and $5^5=125\cdot 25 < 128 \cdot 25 = 32 \cdot (4\cdot 25) = 3200$.) This proves $(**)$.

So we have shown the following chain of inequalities $$\ln \frac54 \ln 4> \frac15 \ln 4 > \frac17\ln5 > \ln\frac87\ln 5.$$ If we compare the leftmost and the rightmost expression, this is precisely $(*)$.

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$\newcommand{\Log}{\operatorname{Log}}$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: $$\Log(3.5)+\frac{2}{7}(x-3.5)-\frac{2}{49}(x-3.5)^2+O(x-3.5)^3$$ For $x=5$ we get: $\Log(5)>33/98+\Log[3.5]$

On the other hand for $x=4$ the linear term gives: $\Log(4)<1/7+\Log(3.5)$. Now we aim to prove the following:$$ \Log(3.5) \Log(5) > \Log(3.5)*(33/98 + \Log(3.5)) > (1/7 + \Log(3.5))^2 > \Log(4)^2$$

The first and last are proved already by the approximations. We need the middle. It is equivalent to the positivity of:$$ \Log(3.5)*(33/98 + \Log(3.5)) - (1/7 + \Log(3.5))^2 =-(1/49) + 5/98 \Log(3.5) =(1/49)(5/2\Log(3.5)-1) $$

The last is positive because $\Log(3.5)>1>2/5$.

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I would have thought you needed $$\frac{\log(5) - \log(4)}{\log(1) - \log(5)}<\frac{\log(8) - \log(7)}{\log(1) - \log(4)}$$ – Henry Sep 20 '12 at 9:00
This is exactly what I prove. Reversing up the first inequality by multiplying the denominators by -1. – ivan Sep 20 '12 at 9:01
My denominators are not the same as your final line – Henry Sep 20 '12 at 9:02
yes, you are right... this proof is wrong – ivan Sep 20 '12 at 9:07
I have completely redone the proof, so the comments above are obsolete. – ivan Sep 20 '12 at 11:42

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