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

Anakin · 2010-09-22 15:31:06

So for some reason, my high school had never gone over this and during the review for Calculus in university, a question came up that I could not solve. Everything seems very straight-forward but not this. Any tips?

For instance, I found this question on Yahoo Answers with a solution but the steps were a bit confusing as it was written using * and ^, etc rather than LaTeX or such. Here is my attempt at the solution:

Thanks for having a look!

Edit: I was able to get the solution by rereading what the Yahoo Answers poster wrote.

My TA was mentioning something about reversing the signs when the base of the logarithmic function is 0<x<1, sort of like how you do so when multiplying both sides by a negative number. What is that all about?

Last edited by Anakin (2010-09-22 16:13:09)

Anakin · 2010-09-22 16:07:01

Here is my understanding of the log inequalities with some curiosities:

Sorry, my HS teacher never went through this stuff for some odd reason but I won't let that stop me!

Last edited by Anakin (2010-09-22 16:14:01)

bobbym · 2010-09-22 16:12:50

Hi Anakin;

Just a thought but you only really need to solve

Anakin · 2010-09-22 16:15:54

Sorry Bobby, I was making a few edits there (taking out some of the stuff that I was able to get).

From what part of the solution did you get the 2^x < 1? The last part with the -2 < 2^x < 1? If so, I was able to understand how to get that part by splitting the inequality into two and solving.

However, the other question remains.

Last edited by Anakin (2010-09-22 16:16:46)

bobbym · 2010-09-22 16:22:28

Hi Anakin;

Maybe because of

Anakin · 2010-09-22 16:27:30

Bobby, could you elaborate a bit with how that affects the inequalities?

(1/3)^-1 = 3
but that's the same as saying 3 = (1/3)^-1

However with the inequalities, that does not work.

Last edited by Anakin (2010-09-22 16:28:00)

bobbym · 2010-09-22 16:34:28

Hi;

I am working with this one which I know the answer to:

The answer is -1 < x < - ( 2 / 3 ). If you sove that you will see that you must switch the inequality to get the x < - ( 2 / 3 )

Anakin · 2010-09-22 16:39:58

That's a good way of understanding it. Would you say that the following generalizations are correct?
logb a > c [assuming that the log part is on the left]
then we keep the a on the left side and get:

a > b^c

If b is 0<x<1:
then we switch the sign and get

a < b^c

bobbym · 2010-09-22 16:46:26

I was thinking maybe for the second one:

0 < a < b^c

What do you think?

I don't like that!

Look at this definition:

Anakin · 2010-09-22 16:58:02

Oops, I meant to say if 0<b<1, not x.

As for 0 < a < b^c, I tried out some examples and it does seem to check out. If I use it when b is equal to 3 or something, it does not work so that means it is right in that aspect (that is works for 0 < b < 1).

Just for confirmation, are you entirely certain that the second part is accurate or are you just testing me to see if I'm actually thinking about this problem? Haha.

Anakin · 2010-09-22 17:00:25

.

Last edited by Anakin (2010-09-22 17:04:41)

bobbym · 2010-09-22 17:01:02

Hi Anakin;

Just for confirmation, are you entirely certain that the second part is accurate or are you just testing me to see if I'm actually thinking about this problem? Haha.

Wish I was cause then I would be sure. I put a new definition up there but it is based on ln(b) ( the natural log ). This one is correct but much harder to understand.

Anakin · 2010-09-22 17:03:03

Repost as my edit was too late: I did not notice your edit. So to reiterate, if C is an element of all real numbers and the log of the base is less than 0, then 0 < a < b^c. But if the log of the base is greater than 0, then a > b^c. Did I understand that correctly?

In essence the log of a base that is between 0 and 1 is always negative and if it is greater than 1, it is positive so the part about 0 < b < 1 and b>1 does stand true, correct?

Last edited by Anakin (2010-09-22 17:03:30)

bobbym · 2010-09-22 17:10:09

Hi Anakin;

I think so, at least that is how I understand it. Remember it is the natural log, ln which is to base e. Please find out from your TA what he thinks is correct and repost his answer here if you can. There might be something cleaner and shorter.

Anakin · 2010-09-22 17:12:35

Alright then. Thank you for all the help Bobby. Even if it may not be as clear as possible (as you put it), it still made sense and cleared up the problem for me.

I'll ask my TA sometime this week and post the answer.

bobbym · 2010-09-22 17:15:19

I am glad it helped a little. Good luck with your studies!

Anakin · 2010-09-22 17:18:18

Thank you!

Anakin · 2010-09-24 16:32:45

Just an update, the TA said something practically along the lines of what you mentioned, Bobby.

Keep the A value on the side that it is on. Solve the inequality of A > 0. Then we solve the WHOLE inequality by isolating A. If B is between 0 and 1, then we simply switch the inequality sign, if not, we leave the sign as it is. Once that inequality is solved, we look at the inequality we solved earlier (A > 0) and conjure a solution by using methods such as a number line or whatever.

Those were her words, which seems quite on par with what you said I think.

Last edited by Anakin (2010-09-24 16:33:29)

bobbym · 2010-09-24 16:38:29

Hi Anakin;

Thanks for providing that. I appreciate it.

sameer mishra · 2010-09-24 17:01:53

dear friend anakin in my openion you do mistake in inquality
~if log b (a)>c
if base of log is {.}
then inequality will change
if>=1 then as it remain

Anakin · 2010-09-25 15:28:03

No prob, Bobby!

And Sameer, what does it mean by saying {.}?

bobbym · 2010-09-25 17:15:55

Hi Anakin;

I think he means > 0 and < 1. Those would have a decimal point up front.

Anakin · 2010-09-25 20:26:10

That is probably it but that would mean he is suggesting precisely what we suggested: a number between 0 and 1. I guess we're all right.

bobbym · 2010-09-25 21:08:03

Hi Anakin;

Yes, I think so.

mohit purswani · 2011-07-20 19:35:41

hi can anyone solve this and give me the steps for this question:-
(x^(log5 x))>5

Math Is Fun Forum

#1 2010-09-22 15:31:06

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

#2 2010-09-22 16:07:01

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

#3 2010-09-22 16:12:50

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

#4 2010-09-22 16:15:54

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

#5 2010-09-22 16:22:28

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

#6 2010-09-22 16:27:30

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

#7 2010-09-22 16:34:28

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

#8 2010-09-22 16:39:58

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

#9 2010-09-22 16:46:26

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

#10 2010-09-22 16:58:02

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#11 2010-09-22 17:00:25

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#12 2010-09-22 17:01:02

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#13 2010-09-22 17:03:03

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#14 2010-09-22 17:10:09

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

#15 2010-09-22 17:12:35

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

#16 2010-09-22 17:15:19

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#17 2010-09-22 17:18:18

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

#18 2010-09-24 16:32:45

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

#19 2010-09-24 16:38:29

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

#20 2010-09-24 17:01:53

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

#21 2010-09-25 15:28:03

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

#22 2010-09-25 17:15:55

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

#23 2010-09-25 20:26:10

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

#24 2010-09-25 21:08:03

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

#25 2011-07-20 19:35:41

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

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