Inequality

great_math · #1 $Old$ 11-05-2008, 05:37 AM

Prove that $\log_7 10>\log_11 13$

Rapha · #2 $Old$ 11-07-2008, 09:56 PM

Quote:

Originally Posted by great_math $View Post$

Prove that $\log_7 10>\log_11 13$

Hi.
I guess you mean $\log_7 (10)>\log_{11} (13)$ , right?

Do you know the formula l $og_a (x) = \frac{ln(x)}{ln(a)}$

Hence

$log_7 (10) = \frac{ln(10)}{ln(7)} > \frac{ln(13)}{ln(11)} = log_{11}(13)$

$\Rightarrow \frac{ln(10)*ln(11)}{ln(7)*ln(13)} > 1$
...

JaneBennet · #3 $Old$ 11-10-2008, 02:29 AM

Quote:

Originally Posted by great_math $View Post$

Prove that $\log_7 10>\log_{\color{red}11 }\color{red}13$

First note that the function $g(x)=x\ln{x}$ is strictly increasing for $x>1.$ (It is in fact strictly increasing for $x>e^{-1}$ but we take $x>1$ for simplicity’s sake.)

Hence for all $1<a<b.$ we have $a\ln{a}<b\ln{b};$ i.e. $\frac{\ln{a}}b-\frac{\ln{b}}a<0\ \ldots\fbox{1}$

Now define $f(x)=\ln(10+x)\ln(10-x)$ for $0<x<9.$

We have $f'(x)=\frac{\ln(10-x)}{10+x}-\frac{\ln(10+x)}{10-x}.$

Since $1<10-x<10+x$ for $0<x<9,$ we have that $f'(x)<0$ for $0<x<9$ by $\fbox{1}$ above.

Thuis $f(x)$ is strictly decreasing for $0<x<9.$

In particular, $f(1)>f(3)$

i.e. $\ln{11}\cdot\ln{9}\ >\ \ln{13}\cdot\ln{7}$

But $\ln{10}>\ln{9}.$

$\therefore\ \ln{11}\cdot\ln{10}\ >\ \ln{13}\cdot\ln{7}$

$\Rightarrow\ \frac{\ln{10}}{\ln{7}}\ >\ \frac{\ln{13}}{\ln{11}}$