Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#26 2011-07-20 20:11:29

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Solving Logarithmic Inequalities With a Base That Is 0<x<1

Hi mohit purswani;

Welcome to the forum!

This is what I tried.

Take the natural log of both sides:

Times both sides by ln(5):

Take e^ of both sides:

We reject x < - 5 as not solving the original inequality. So we have x = (5, ∞)

The "Take e^ of both sides:"  has knocked out some solutions. There are more solutions in ( 0 , 1 / 5 ) but I can not prove that.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Offline

#27 2011-07-20 21:36:33

Bob
Administrator
Registered: 2010-06-20
Posts: 10,621

Re: Solving Logarithmic Inequalities With a Base That Is 0<x<1

hi Anakin and bobbym

re: Anakin's OP.

It's because the graph of y = logx in base b where b is fractional is decreasing.

This is true for any 'b'.

But now let b = 1/2         =>       ln(b) = -0.693147180559945

Now see my graph below.

The 'y' value decreases as 'x' increases.

So as you change from the log equation to the power equation you must reverse the inequality.

This is true for any decreasing graph.

eg.  60 > 30 but cos(60) < cos(30)  angles in degrees.

Bob

Last edited by Bob (2011-07-20 21:37:49)


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

Offline

Board footer

Powered by FluxBB