Again a logarithmic inequality

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Homework Help Overview

The discussion revolves around a logarithmic inequality involving expressions with logarithms and algebraic manipulation. Participants are exploring the relationships between different logarithmic terms and their simplifications.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss simplifying logarithmic expressions and the implications of those simplifications. Questions arise regarding the differences between various logarithmic forms and how they relate to the original inequality.

Discussion Status

There are multiple lines of inquiry, with participants suggesting different methods such as graphing and solving associated equations. Some guidance has been offered regarding simplifications and the exploration of critical points, but no consensus has been reached on a specific approach.

Contextual Notes

Participants note the need to find values of x, and there is mention of points of discontinuity in the context of the inequality. The discussion includes references to specific logarithmic properties and expressions that may require further clarification.

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Homework Statement


i got stuck at the question below:-

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Homework Equations




The Attempt at a Solution


I tried to solve it by simplifying it but i got stuck at:-

250t2zb.png


Please help.
 
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Hi Pranav-Arora! :smile:

Just simplify the bottom …

what is the difference between log3(9 - 3x) and log3(1 - 3x-2) ? :wink:
 
tiny-tim said:
Hi Pranav-Arora! :smile:

Just simplify the bottom …

what is the difference between log3(9 - 3x) and log3(1 - 3x-2) ? :wink:

Sorry! Didn't get you...
 
One method for solving an inequality is to solve the associated equation; in this case that's

[tex]\frac{x-1}{\log_3(9-3^x)-3}=1[/tex].

Then the critical numbers are the solution set and any points of discontinuity.

Use test points (in the domain of the left hand side of the inequality) which either to the laeft or right of all the test points or between any pair of test point.

By the way, what is log3(9(1-3x-2)) ?
 
SammyS said:
One method for solving an inequality is to solve the associated equation; in this case that's

[tex]\frac{x-1}{\log_3(9-3^x)-3}=1[/tex].

Then the critical numbers are the solution set and any points of discontinuity.

Use test points (in the domain of the left hand side of the inequality) which either to the laeft or right of all the test points or between any pair of test point.

By the way, what is log3(9(1-3x-2)) ?

i think i forgot to mention, i need to find out the values of x.
 
Pranav-Arora said:
Sorry! Didn't get you...

Hi Pranav-Arora! :smile:

what is log3(9 - 3x) - log3(1 - 3x-2) ? :wink:
 
tiny-tim said:
Hi Pranav-Arora! :smile:

what is log3(9 - 3x) - log3(1 - 3x-2) ? :wink:

It would be

9 - 3x
-------
1 - 3x-2

But why i need to find this?
 
What is

9 - 3x
-------
1 - 3x-2

? :wink:
 
Try graphing [tex]\frac{x-1}{\log_3(9-3^x)-3}[/tex] or [tex]\frac{x-1}{\log_3(9-3^x)-3}-1\,.[/tex]

Remember that [tex]\log_3(a)=\frac{\ln(a)}{\ln(3)}[/tex]
 
  • #10
SammyS said:
Try graphing [tex]\frac{x-1}{\log_3(9-3^x)-3}[/tex] or [tex]\frac{x-1}{\log_3(9-3^x)-3}-1\,.[/tex]

Remember that [tex]\log_3(a)=\frac{\ln(a)}{\ln(3)}[/tex]

Whoops! i forgot it:-
[tex]\log_3(a)=\frac{\ln(a)}{\ln(3)}[/tex]

I converted everything in log and then i was able to figure it out.

Thanks SammyS...:)
 

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