Solve ln (3+x)=7: Get x=1093.63

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Homework Help Overview

The discussion revolves around solving the equation involving a natural logarithm, specifically ln(3+x)=7, with a focus on finding the value of x. Participants are examining the implications of approximating the solution versus expressing it in exact form.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to solve the equation and arrives at an approximate value for x. Some participants question the necessity of providing an exact answer instead of an approximation, suggesting that the method used was valid.

Discussion Status

The discussion is active, with participants providing feedback on the original poster's approach. There is a recognition of the validity of the method used, along with suggestions for presenting the answer in a different form. No explicit consensus has been reached regarding the preferred format of the answer.

Contextual Notes

Participants are considering the implications of approximating the solution and the potential for expressing it in a more exact manner. There is an underlying assumption that the original poster's method is fundamentally sound, despite the concern over the approximation.

jacy
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Hello,
Am trying to solve this natural log problem, where i have to find the value of x.

ln (3+x)=7

this is what i have done

3+x= e7
x= 1093.63

This is the answer am getting, but its wrong. Can someone please let me know what am doing wrong, thanks.
 
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Well the answer isn't 'wrong', but it's only an approximation, why not leave it as an exact answer? It seems that your method was ok...

[tex]\ln \left( {3 + x} \right) = 7 \Leftrightarrow 3 + x = e^7 \Leftrightarrow x = e^7 - 3 \Leftrightarrow x \approx 1093.633158 \ldots[/tex]

So instead of doing that last stap (see how I didn't the equality sign?) just leave it as [itex]x = e^7 - 3[/itex].
 
TD said:
Well the answer isn't 'wrong', but it's only an approximation, why not leave it as an exact answer? It seems that your method was ok...

[tex]\ln \left( {3 + x} \right) = 7 \Leftrightarrow 3 + x = e^7 \Leftrightarrow x = e^7 - 3 \Leftrightarrow x \approx 1093.633158 \ldots[/tex]

So instead of doing that last stap (see how I didn't the equality sign?) just leave it as [itex]x = e^7 - 3[/itex].

Thanks for ur help.
 
You're welcome :smile:
 

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