- #1
Jaco Viljoen
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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)-e^x-2=0I don't know what to do here:
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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.Jaco Viljoen said: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:
Jaco Viljoen said: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.
What you wrote in post #1 should be something of a hint.Jaco Viljoen said:Hi Mark,
Thank you,
I initially did this but how do I factor e^(x2)
Jaco Viljoen said:e^(2x)-2=e^x
((e^x)^2)-e^x-2=0
If you have terms added together, taking the log is not any help.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.
This is correct, but you should say why your are discarding the other value of k.Jaco Viljoen said:((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
The OP has already gotten all of this.NicolasPan said:1.a) Dh=ℝ
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 function h(x)=e^(2x)-2 is an exponential function that involves the mathematical constant e raised to the power of 2x, with the result then being subtracted by 2. It is also known as the natural exponential function.
The domain of the function h(x)=e^(2x)-2 is all real numbers. This means that any value of x can be plugged into the function and have a valid output.
The range of the function h(x)=e^(2x)-2 is all real numbers greater than or equal to -2. This is because the exponential function can only output positive values, and the subtraction of 2 shifts the range downwards.
The y-intercept of the function h(x)=e^(2x)-2 is -1. This can be found by setting x=0 and solving for y. The resulting point would be (0, -1).
The graph of h(x)=e^(2x)-2 is the graph of e^(2x) shifted down by 2 units. This means that the two graphs have the same shape, but one is shifted downwards. Additionally, the y-intercept of h(x) is -1, while the y-intercept of e^(2x) is 1.