Integration - Find the x coordinate and the area under the curve

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Homework Help Overview

The discussion revolves around finding the x-coordinate where two exponential functions intersect and calculating the area under the curve defined by these functions. The subject area includes integration and logarithmic properties.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to set the two equations equal to each other and take logarithms to solve for x. Some participants question the handling of the logarithmic properties and the implications of fractions in the equations.

Discussion Status

Participants are actively engaging with the original poster's attempts, providing hints and clarifications about logarithmic manipulation. There is a focus on ensuring the correct application of logarithmic rules, but no consensus has been reached on the final solution.

Contextual Notes

The original poster expresses uncertainty about the logarithmic steps and how to express the final answer in a specific form. There are indications of potential misunderstandings regarding the properties of logarithms and fractions in the equations.

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Homework Statement



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Homework Equations



To find the x coordinate

1. Make both equations equal, expose e and take logs. I'm not sure how to do this and I've tried but keep getting the wrong answer.

2. To find the area, subract the higher curve from the lower curve and integrate. I can do this. I only need help finding the x coordinate.

The Attempt at a Solution



1. y=e^(x-3)/2 and y =e^(2x-7)

e^(x-3)/2 =e^(2x-7)

Now take logs of both sides. How?
 
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[tex]\ln (e^{f(x)})=f(x)[/tex] But watch out for the fraction over 2. First, move it over to the other side. Then be careful, as [itex]\ln (2e^{2x-7}) \neq 2e^{2x-7}[/itex]
 
ArcanaNoir said:
[tex]\ln (e^{f(x)})=f(x)[/tex] But watch out for the fraction over 2. First, move it over to the other side. Then be careful, as [itex]\ln (2e^{2x-7}) \neq 2e^{2x-7}[/itex]


Thanks for your reply. I get:

1/2e^(x-3)=e^(2x-7)
e^(x-3)=2(e^(2x-7))
lne^(x-3)=ln(e^(2x-7))
(x-3)/2=2x-7
x=11/3

'Express your answer in the form a-lnb.How?
 
Whoops! your 2 disappeared for a line, then it came back. You can't do that. You still need to figure out what [itex]\ln (2e^{2x-7})[/itex] is. Hint: [itex]\ln (a) + \ln (b)=\ln (ab)[/itex]
 

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