Identities for solving log questions

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SUMMARY

The discussion centers on the use of logarithmic identities to solve equations involving logarithms, specifically expressions like (loga b)² and (loga b)(loga c). Participants clarify that while there are no specific identities for these forms, simplification can be achieved through substitution and algebraic manipulation. For example, substituting y = loge(x) allows for solving (loge x)² effectively. The conversation emphasizes the importance of understanding the properties of logarithms and the behavior of inequalities in logarithmic equations.

PREREQUISITES
  • Understanding of logarithmic properties and identities
  • Familiarity with algebraic manipulation techniques
  • Knowledge of substitution methods in equations
  • Basic understanding of inequalities and their properties
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  • Study logarithmic properties in depth, focusing on common identities
  • Practice solving logarithmic equations using substitution methods
  • Learn about inequalities involving logarithmic functions
  • Explore advanced logarithmic applications in calculus and algebra
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Students, educators, and anyone seeking to improve their skills in solving logarithmic equations and understanding logarithmic identities.

  • #91
Hi Pranav-Arora! :smile:

Pranav-Arora said:
Thanks!It worked! :smile:

My next question:-

\log_3\frac{|x^2-4x|+3}{x^2+|x-5|} \ge 0

Hmm, that's not a question - it's a problem.
What is your question? :wink:

I suspect that the log should bother you no more by now, nor the matter of the domain of the log function.

The only new thing is the absolute value function...
And I'm not sure whether you fully understand when an inequality inverts and how exactly...

Perhaps show a few steps...?
 
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  • #92
I like Serena said:
Hi Pranav-Arora! :smile:



Hmm, that's not a question - it's a problem.
What is your question? :wink:

I suspect that the log should bother you no more by now, nor the matter of the domain of the log function.

The only new thing is the absolute value function...
And I'm not sure whether you fully understand when an inequality inverts and how exactly...

Perhaps show a few steps...?

When i tried to solve it, i wasn't able to go further than this:-

\frac{|x^2-4x|+3}{x^2+|x-5|} \ge 1

I am not able to figure out what should i do next? :confused:
 
  • #93
Pranav-Arora said:
When i tried to solve it, i wasn't able to go further than this:-

\frac{|x^2-4x|+3}{x^2+|x-5|} \ge 1

I am not able to figure out what should i do next? :confused:

What would happen if you multiplied left and right by the denominator?

Can you separate this in a number of cases for x?
That is, for instance, what would the signs of the absolute values be if x > 5?
For which values of x would one or the other absolute value invert its sign?
 
  • #94
Pranav-Arora said:
When i tried to solve it, i wasn't able to go further than this:-

\frac{|x^2-4x|+3}{x^2+|x-5|} \ge 1

I am not able to figure out what should i do next? :confused:

What would happen if you multiplied left and right by the denominator?

Can you separate this in a number of cases for x?
That is, for instance, what would the signs of the absolute values be if x > 5?
For which values of x would one or the other absolute value invert its sign?
 
  • #95
I like Serena said:
What would happen if you multiplied left and right by the denominator?

Can you separate this in a number of cases for x?
That is, for instance, what would the signs of the absolute values be if x > 5?
For which values of x would one or the other absolute value invert its sign?

Absolute values are positive only...so why are you asking me "what would the signs of the absolute values be if x > 5?" :confused:
 
  • #96
Pranav-Arora said:
Absolute values are positive only...so why are you asking me "what would the signs of the absolute values be if x > 5?" :confused:

What I meant is that you can write an absolute value without the vertical bars if you put the appropriate sign before it in combination with a domain restriction.

For instance:
|x-5| = (x-5) if x > 5
|x-5| = -(x-5) if x < 5
 
  • #97
Look at the denominator. Can it be negative?

ehild
 

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