The Function h(x)=e^(2x)-2

  • #1
160
9

Homework Statement


h(x)=e^(2x)-2
a)Write down the Dh
b)Solve the equation e^(2x)-2=e^x

Homework Equations




The Attempt at a Solution


a)x∈ℝ[/B]
e^(2x)-2=e^x
((e^x)^2)-e^x-2=0


I don't know what to do here:
 
Last edited:

Answers and Replies

  • #2
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Homework Statement


h(x)=e^(2x)-2
a)Write down the Dh
b)Solve the equation e^(2x)-2=e^x

Homework Equations




The Attempt at a Solution


a)x∈ℝ[/B]
e^(2x)-2=e^x
((e^x)^2)-2=e^x

I don't know what to do here:
Move the ex term to the left side. Your equation is quadratic in form, and can be factored. Keep in mind that ex > 0 for all x.
 
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  • #3
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Hi Mark,
Thank you,
I initially did this but how do I factor e^(x2)
I also considered adding a ln to both sides to get rid of the e on both sides but the -2 is the problem.
 
  • #4
Ray Vickson
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Hi Mark,
Thank you,
I initially did this but how do I factor e^(x2)
I also considered adding a ln to both sides to get rid of the e on both sides but the -2 is the problem.
You need to go back and review the laws of exponents. How does ##e^{2x}## relate to ##e^x##? Surely you must have had such material already, or else you would not have been asked to do this problem.
 
  • #5
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6,193
Hi Mark,
Thank you,
I initially did this but how do I factor e^(x2)
What you wrote in post #1 should be something of a hint.
Jaco Viljoen said:
e^(2x)-2=e^x
((e^x)^2)-e^x-2=0
Jaco Viljoen said:
I also considered adding a ln to both sides to get rid of the e on both sides but the -2 is the problem.
If you have terms added together, taking the log is not any help.
 
  • #6
160
9
((e^x)^2)-2=e^x
(k^2)-k-2=0 k=e^x
(k+1)(k-2)=0
k=-1 or k=+2

e^x=2
lne^x=ln2
x=ln2
 
  • #7
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6,193
((e^x)^2)-2=e^x
(k^2)-k-2=0 k=e^x
(k+1)(k-2)=0
k=-1 or k=+2

e^x=2
lne^x=ln2
x=ln2
This is correct, but you should say why your are discarding the other value of k.
 
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  • #8
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1.a) Dh=
b)The solution can go through this path: (e^x)^2-e^x-2=0
If e^x=y i) then the equation is transformed like that: y^2-y-2=0
then the solutions are y=(1±3)/2 and from i) we get e^x=(1±3)/2 BUT since the function e^x can only give positive results for x∈ℝ the only acceptable solution would be the one with the +. So e^x=2 ⇔ x=ln2, there you go friend I hope this helps :)
 
  • #9
34,509
6,193
1.a) Dh=
b)The solution can go through this path: (e^x)^2-e^x-2=0
If e^x=y i) then the equation is transformed like that: y^2-y-2=0
then the solutions are y=(1±3)/2 and from i) we get e^x=(1±3)/2 BUT since the function e^x can only give positive results for x∈ℝ the only acceptable solution would be the one with the +. So e^x=2 ⇔ x=ln2, there you go friend I hope this helps :)
The OP has already gotten all of this.
 

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