Solve the following equation ln(x^2-8x+13)=0

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Homework Help Overview

The problem involves solving the equation ln(x^2 - 8x + 13) = 0, which is situated within the context of logarithmic functions and their properties.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to manipulate the logarithmic equation but expresses uncertainty about their direction. Some participants clarify the properties of logarithms, particularly the distinction between ln(a + b) and ln(ab), and suggest taking the exponential of both sides.

Discussion Status

Participants are engaging with the problem, with some providing clarifications on logarithmic properties. There is an acknowledgment of the need to consider the positivity of the argument within the logarithm, indicating a productive exploration of the topic.

Contextual Notes

The original poster mentions being new to the material, which may influence their understanding and approach to the problem. There is also a note about the necessity to discard values that do not meet the conditions for logarithmic functions.

A_Munk3y
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Homework Statement


Solve the following equation:
ln(x2-8x+13)=0



The Attempt at a Solution


not much luck with this one. i just learned this stuff and I am having problems with it'
Anyways, i got
ln(x2-8x+13=0
=>lnx2-ln8x+ln13=0
=>lnx2-ln8x=-ln13
=>ln(x2/8x)=-ln13
=>eln(x2/8x)=e-ln13
=>x2/8x=e-ln13

im just going to stop there cause i think I am going in the wrong direction
 
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[tex]ln(a+b)\neq ln(a)+ln(b)[/tex]

But rather,

[tex]ln(ab)=ln(a)+ln(b)[/tex]

Firstly, take the exponential of both sides, since [tex]e^{ln(x)}=x[/tex]

But also remember that you can only take the log of a positive number, so you might need to discard a value when you solve for x.
 
Oh, ok :)
I got it! Thank you
 
No problem :smile:
 

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