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Solving a logarithmic equation

  • Thread starter xxwinexx
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  • #1
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Homework Statement


(e^x)^2-5(e^x)=0

Homework Equations


I've reviewed my logarithmic rules, but I cannot get to the right solution. I should be trying to get one e^x on each side so that I can take the ln of both sides and end up with a simple algebraic process to solve.


The Attempt at a Solution


This is as far as I get without running into my problem:
(e^x)^2=5(e^x)

I guess here I could take the ln of both sides, but I guess I'm confused about what happens to the x^2, and to the 5.

Edit:

Ahh, Thanks for the reminder. Don't wanna bump this any more.

Took ln of both sides, ended with:

2x=ln5+x

x=ln5

Is that the correct way to that solution?
 
Last edited:

Answers and Replies

  • #2
1,013
65
Remember your exponent and logarithm rules (they're pretty similar): If a, b, and c are real numbers, then:
(ab)c = abc
and
ln(a) + ln(b) = ln(ab).
The latter is derived from the fact that eaeb = ea+b.
 
  • #3
1,013
65
Indeed. You can check it by replacing x with ln(5) in the original equation. Since both sides are positive (the natural logarithm has only a positive domain), it is also the only real solution.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
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Homework Statement


(e^x)^2-5(e^x)=0

Homework Equations


I've reviewed my logarithmic rules, but I cannot get to the right solution. I should be trying to get one e^x on each side so that I can take the ln of both sides and end up with a simple algebraic process to solve.
If you let y= ex, then your equation is y2- 5y= y(y- 5)= 0. Can you solve that?


The Attempt at a Solution


This is as far as I get without running into my problem:
(e^x)^2=5(e^x)

I guess here I could take the ln of both sides, but I guess I'm confused about what happens to the x^2, and to the 5.
?? There is NO x^2. There is (e^x)^2.

Edit:

Ahh, Thanks for the reminder. Don't wanna bump this any more.

Took ln of both sides, ended with:

2x=ln5+x

x=ln5

Is that the correct way to that solution?
 

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