So my teachers notes say: 2^(2x) = 2^(2x) / 2Ln(2)
why is it 2 times ln(2) instead of just ln(2)?
thank you
So my teachers notes say: 2^(2x) = 2^(2x) / 2Ln(2)
why is it 2 times ln(2) instead of just ln(2)?
thank you
So my teachers notes say: 2^(2x) = 2^(2x) / 2Ln(2)
why is it 2 times ln(2) instead of just ln(2)?
What happens when you try integrating it?
There are a couple ways you might be approaching it. Do you have a formula for the integral of, say, a^u du? Or are you starting with the formula for the integral of e^u du and rewriting your integrand in terms of that? Either way, you will need to use substitution; and that is where the 2 comes from. The ln(2) comes from the fact that the base is 2.
But you won't fully grasp what is happening without doing it for yourself. So give it a try!
Write this function in the form of an exponential to base e instead of base 2, so that you know how to differentiate it and integrate it. First take the ln of the function
[tex] \displaystyle \ln\left(2^{2x}\right) = 2x\ln(2) [/tex]
Now undo this by raising e to the power of both sides. Since exponentiating and taking the natural log (i.e. the base e log) are just inverse operations, these two steps should undo each other, and you get back what you started with on the lefthand side:
[tex] \displaystyle e^{\ln\left(2^{2x}\right)} = e^{2x\ln(2)} [/tex]
[tex] \displaystyle 2^{2x} = e^{2x\ln(2)} [/tex]
So we've expressed the function as a power with base e, instead of base 2. I hope that makes perfect sense. If it doesn't, ask now, because you need to be able to do the steps on your own in an exam. Now we can compute your integral by integrating both sides of the above equation:
[tex] \displaystyle \int 2^{2x}\,dx = \int e^{x(2\ln(2))}\,dx [/tex]
The integral on the righthand side is one that you should know how to do, because it is just of the form [tex]\int e^{ax} \,dx[/tex] where [tex]a[/tex] is some constant. Can you take it from here?
Q: What happens when you post the same question in multiple places, on multiple bulletin boards, and you get answers from multiple volunteers?
A: You are declaring that your time is more valuable than the time of the capable volunteers.
Please rethink this behavior. If you must have an immediate answer, or perhaps your "exam is tomorrow," you might consider paying for it. Volunteers come by when they can. We don't really do urgency, except as a coincidence.
"Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.
There is a duplicate of this thread in Intermediate/Advanced Algebra that I responded to:
https://www.freemathhelp.com/forum/threads/110418-what-is-the-integral-of-2-(2x)-need-help-tonight-exam-is-tomorrow
EDIT: I just realized that tkhunny already figured that out. OK
The other thread here for this question has now been moved from "Algebra" to "Calculus".
If the original poster thought that this was an "Algebra" question, this could explain the lack of work or response.
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