Use the properties of logarithma to xpand the logarithmic function

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Homework Help Overview

The discussion revolves around expanding the logarithmic function ln[(x²+1)(x-1)] using properties of logarithms. Participants are exploring the correct application of logarithmic identities in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the initial step of breaking down the logarithm into separate terms, questioning the validity of further expansion of ln(x²+1). There is consideration of factoring and the implications of complex numbers in the context of x²+1.

Discussion Status

The conversation is ongoing, with participants examining different interpretations of how to handle the logarithmic expression. Some guidance has been offered regarding the limitations of factoring certain terms, but no consensus has been reached on the next steps.

Contextual Notes

There is a recognition that the term x²+1 cannot be factored in the realm of real numbers without involving imaginary numbers, which adds complexity to the problem.

Ki-nana18
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Homework Statement


Use the properties of logarithma to xpand the logarithmic function ln[(x2+1)(x-1)]


Homework Equations





The Attempt at a Solution


[ln x2+ln 1]+[ln x-ln 1]
2 ln x+ln x
 
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the first step makes sense: ln[(x^2+1)(x-1)]=ln(x^2+1)+ln(x-1)

but then you continued: ln(x^2+1)=ln(x^2)+ln(1)

You can't do that, but you can do something else to the x^2+1...
 
gamer_x_ said:
the first step makes sense: ln[(x^2+1)(x-1)]=ln(x^2+1)+ln(x-1)

but then you continued: ln(x^2+1)=ln(x^2)+ln(1)

You can't do that, but you can do something else to the x^2+1...
Is the something else you're thinking about factoring x^2 + 1?
 
I just realized I was stupidly thinking of x^2-1 not x^2+1. You can't really factor that term unless you go into imaginary numbers.
 
That's what I thought you might be thinking.
 

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