# Logarithmic inequality

• Oct 19th 2008, 05:51 AM
achacy
Logarithmic inequality
$\displaystyle log _{ \frac{1}{3} } x> log _{x}3-2,5$

Most important for me is the result as I made this exercise and my result is different as given in the book.
• Oct 19th 2008, 07:12 AM
Soroban
Hello, achacy!

I have a somewhat intricate solution.
. . Maybe someone can do better . . .

Quote:

$\displaystyle \log _{\frac{1}{3}}x \;>\; \log _{x}3 - 2.5$ .[1]

I converted the terms to the same base.

$\displaystyle \text{Let: }\log_{\frac{1}{3}}(x) \:=\:P \quad\Rightarrow\quad \left(\frac{1}{3}\right)^P =\:x$

. . Take logs: .$\displaystyle \ln\left(\frac{1}{3}\right)^P =\:\ln(x) \quad\Rightarrow\quad P\ln\left(\frac{1}{3}\right) \:=\:\ln(x)$

. . $\displaystyle P\ln\left(3^{-1}\right) \:=\:\ln(x) \quad\Rightarrow\quad-P\ln(3) \:=\:\ln(x) \quad\Rightarrow\quad P \:=\:-\frac{\ln(x)}{\ln(3)}$

. . Hence: .$\displaystyle \log_{\frac{1}{3}}(x) \;=\;-\frac{\ln(x)}{\ln(3)}$ .[2]

From the Base-Change Formula: .$\displaystyle \log_x(3) \:=\:\frac{\ln(3)}{\ln(x)}$ .[3]

Substitute [2] and [3] into [1]: .$\displaystyle -\frac{\ln(x)}{\ln(3)} \;> \;\frac{\ln(3)}{\ln(x)} - 2.5$

We have: .$\displaystyle -\frac{\ln(x)}{\ln(3)} - \frac{\ln(3)}{\ln(x)} + 2.5 \;>\;0 \quad\Rightarrow\quad \frac{\ln(x)}{\ln(3)} + \frac{\ln(3)}{\ln(x)} - 2.5 \;<\;0$

Let $\displaystyle u \:=\:\frac{\ln(x)}{\ln(3)} \quad\Rightarrow\quad u + \frac{1}{u} - 2.5 \;<\;0$

Muliply by $\displaystyle u\!:\;\;u^2 + 1 - 2.5u \;<\;0$ . . . assuming $\displaystyle u > 0\;\;(x > 1)$

Multiply by 2: .$\displaystyle 2u^2 - 5u + 2 \;<\;0\quad\Rightarrow\quad (u-2)(2u-1) \;<\;0$

There are two cases:

. . $\displaystyle \begin{array}{ccc}u-2\:>\:0 & \Rightarrow & u \:>\:2 \\ 2u-1 \:<\:0 & \Rightarrow & u \:<\:\frac{1}{2}\end{array}\quad\hdots \text{ impossible}$

. . $\displaystyle \begin{array}{ccc}u-2 \:<\:0 & \Rightarrow & u \:<\:2 \\ 2u-1 \:>\:0 & \Rightarrow & u \:>\:\frac{1}{2}\end{array}\quad \hdots\;\tfrac{1}{2}\:<\:u\:<\:2$

Then we have: . $\displaystyle \tfrac{1}{2}\;<\:\frac{\ln(x)}{\ln(3)} \;<\;2$

Multiply by $\displaystyle \ln(3)\!:\;\;\tfrac{1}{2}\ln(3) \;<\;\ln(x) \;<\;2\ln(3) \quad\Rightarrow\quad \ln\left(3^{\frac{1}{2}}\right) \;<\;\ln(x) \;<\;\ln\left(3^2\right)$

Therefore: .$\displaystyle \boxed{\sqrt{3}\;<\;x\;<\;9}$

• Oct 19th 2008, 07:26 AM
sosgf
put y = lnx/ln3 ===> -y > 1/y - 5/2

2y^2 -5y + 2 < 0 ===> 1/2< y < 2 ===> rotsq( 3) < x < 9