what is the integral of 2^(2x)? need help tonight, exam is tomorrow

davesterrr

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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
 
what is the integral of 2^(2x)? need help tonight, exam is tomorrow.

So my teachers notes say: 2^(2x) = 2^(2x) / 2Ln(2)

why is it 2 times ln(2) instead of just ln(2)?
 
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

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!
 
So my teachers notes say: 2^(2x) = 2^(2x) / 2Ln(2)

why is it 2 times ln(2) instead of just ln(2)?

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

\(\displaystyle \displaystyle \ln\left(2^{2x}\right) = 2x\ln(2) \)

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:

\(\displaystyle \displaystyle e^{\ln\left(2^{2x}\right)} = e^{2x\ln(2)} \)

\(\displaystyle \displaystyle 2^{2x} = e^{2x\ln(2)} \)

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:

\(\displaystyle \displaystyle \int 2^{2x}\,dx = \int e^{x(2\ln(2))}\,dx \)

The integral on the righthand side is one that you should know how to do, because it is just of the form \(\displaystyle \int e^{ax} \,dx\) where \(\displaystyle a\) is some constant. Can you take it from here?
 
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

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.
 
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. :(
 
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
NO! Maybe the derivative of 2^(2x) = 2^(2x) / 2Ln(2) but 2^(2x) \(\displaystyle \neq\) 2^(2x) / 2Ln(2)
 
soln?

First take the ln of the function
\(\displaystyle \displaystyle \ln\left(2^{2x}\right) = 2x\ln(2) \)
Now undo this by raising e to the power of both sides.
\(\displaystyle \displaystyle e^{\ln\left(2^{2x}\right)} = e^{2x\ln(2)} \)
\(\displaystyle \displaystyle 2^{2x} = e^{2x\ln(2)} \)
Now we can compute your integral by integrating both sides of the above equation:
\(\displaystyle \displaystyle \int 2^{2x}\,dx = \int e^{2x\ln(2))}\,dx \)
The integral on the righthand side is one that you should know how to do, because it is just of the form \(\displaystyle \int e^{ax} \,dx\) where \(\displaystyle a\) is some constant. Can you take it from here?

j-astron; my soln is (1/2)e2x+2x+C
is this correct?
[I am having to re-learn ex, lnx]
 
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j-astron; my soln is (1/2)e2x+2x+C
is this correct?
[I am having to re-learn ex, lnx]
You can check it yourself. Just take the derivative of (1/2)e2x+2x+C and see if you get back the intergrand (the function you were supposed to intergrate)
 
You can check it yourself. Just take the derivative of (1/2)e2x+2x+C and see if you get back the intergrand (the function you were supposed to intergrate)


it is weird, because i figured this out putting y'=22x into form y'=e2xln2 and integrating, but i am having trouble going backwards, i.e. taking the derivative of y=1/2e2x+2x,and putting it into the form 22x.
however, i think i got it;
y=1/2e2x+2x
y'=e2x+2
this should be the integrand, but the trouble is getting e2x+2 into the form 22x

is the following correct?
1st take the ln of both sides of the equation y'=e2x+2;
lny'=2x+ln2;
now reverse this raising both sides to base e;
elny'=e2x+eln2=e2xln2=eln2^2x=22x
y'=22x
​??
 
it is weird, because i figured this out putting y'=22x into form y'=e2xln2 and integrating, but i am having trouble going backwards, i.e. taking the derivative of y=1/2e2x+2x,and putting it into the form 22x.
however, i think i got it;
y=1/2e2x+2x
y'=e2x+2
this should be the integrand, but the trouble is getting e2x+2 into the form 22x

is the following correct?
1st take the ln of both sides of the equation y'=e2x+2;
lny'=2x+ln2;
now reverse this raising both sides to base e;
elny'=e2x+eln2=e2xln2=eln2^2x=22x
y'=22x
​??
Well maybe e2x+2 \(\displaystyle \neq\) 22x. Just let x=0 and you'll see that e2x+2= 3 while 22x=1. For the record I never said that the posted answer was correct. What I did say was that this result can be checked to see if it is correct.
 
Well maybe e2x+2 \(\displaystyle \neq\) 22x. Just let x=0 and you'll see that e2x+2= 3 while 22x=1. For the record I never said that the posted answer was correct. What I did say was that this result can be checked to see if it is correct.

for y=int e2xln2dx
i got y=22x/2ln2
i.e. using,
u=(2ln2)x
du=2ln2dx
then,
y=[1/2ln2]inteudu = ​[1/2ln2]e2xln2 =[1/2ln2]eln2^2x =22x/2ln2

is this correct?,
and i am also having trouble with the derivative of this function to check.

since 2ln2 is a constant, how do i take the derivative (dy/dx) of 22x ? [is it like e2x, where the deriv is 2e2x ?]
alternately, I have tried by making an exponential of both num and denom to base e.
this yields y=e2^2x/e2ln2, =e2^2x/4, stuck

I then tried taking the ln of both sides, lny'=ln [22x/2ln2] = ln22x-ln[2ln2]
then, raising back to e, y'=22x-2ln2. This is incorrect (at least y' is not= 22x​, so either the integral or the derivative are wrong),
then i tried taking log2 of both sides log2y'=log2​[22x/2ln2] = 2x-log22ln2 =2x-log2ln2
and since ln2=1/log2e
log2y'=2x-ln2=2x-log2[log2e]
y'=22x-log2e stuck again.

what am i missing?
 
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for y=int e2xln2dx
i got y=22x/2ln2
i.e. using,
u=(2ln2)x
du=2ln2dx
then,
y=[1/2ln2]inteudu = ​[1/2ln2]e2xln2 =[1/2ln2]eln2^2x =22x/2ln2

is this correct?,
and i am also having trouble with the derivative of this function to check.

since 2ln2 is a constant, how do i take the derivative (dy/dx) of 22x ? [is it like e2x, where the deriv is 2e2x ?]
alternately, I have tried by making an exponential of both num and denom to base e.
this yields y=e2^2x/e2ln2, =e2^2x/4, stuck

I then tried taking the ln of both sides, lny'=ln [22x/2ln2] = ln22x-ln[2ln2]
then, raising back to e, y'=22x-2ln2. This is incorrect (at least y' is not= 22x​, so either the integral or the derivative are wrong),
then i tried taking log2 of both sides log2y'=log2​[22x/2ln2] = 2x-log22ln2 =2x-log2ln2
and since ln2=1/log2e
log2y'=2x-ln2=2x-log2[log2e]
y'=22x-log2e stuck again.

what am i missing?
Yes, y=22x/2ln2 + C

Yes,(dy/dx)(22x) = 2e2x

ey= e2^2x/e2ln2 =e2^2x/4--No! If f(x) = g(x)/h(x), then ef(x) = eg(x)/h(x) \(\displaystyle \neq\) eg(x)/eh(x). What made you think that was true?!
While elnA - elnB = A-B, it is not the case that
elnA-lnB = A-B. In fact, elnA-lnB= eln(A/B)= A/B NOT A-B!!!!!!
 
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Yes, y=22x/2ln2 + C

Yes,(dy/dx)(22x) = 2e2x

ey= e2^2x/e2ln2 =e2^2x/4--No! If f(x) = g(x)/h(x), then ef(x) = eg(x)/h(x) \(\displaystyle \neq\) eg(x)/eh(x). What made you think that was true?!

I was not saying
ey= e2^2x/e2ln2 ; i was saying y=e2^2x/e2ln2 , [i.e. putting both the numerator and denominator of the fraction as exponents in an attempt to make the fraction equal itself, but I can see now this is wrong also. [I was grasping at straws.]

While elnA - elnB = A-B, it is not the case that
elnA-lnB = A-B. In fact, elnA-lnB= eln(A/B)= A/B NOT A-B!!!!!!

how is this?
y'=22x
lny'=2xln2
y'=e2x+eln2 = e2x+2
y=1/2e2x+2x+C

y'=e2x+2
y'=e2x+eln2 = e2xln2
lny'=2xln2=ln22x
y'=22x
 
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how is this?
y'=22x
lny'=2xln2
y'=e2x+eln2 = e2x+2 <--- no, this would have to be y'=e2xln2, which is not e2x+eln2
y=1/2e2x+2x+C

y'=e2x+2
y'=e2x+eln2 = e2xln2
lny'=2xln2=ln22x
y'=22x

You have to use the logarithm rules carefully. One says that a^(m + n) = a^m * a^n, not a^(mn) = a^m + a^n.
 
Yes, y=22x/2ln2 + C

Yes,(dy/dx)(22x) = 2e2x

ey= e2^2x/e2ln2 =e2^2x/4--No! If f(x) = g(x)/h(x), then ef(x) = eg(x)/h(x) \(\displaystyle \neq\) eg(x)/eh(x). What made you think that was true?!
While elnA - elnB = A-B, it is not the case that
elnA-lnB = A-B. In fact, elnA-lnB= eln(A/B)= A/B NOT A-B!!!!!!

I believe I found my mistake;
Yes, y=22x/2ln2 + C,
I take it you mean this is the correct answer.
i got here this way
y'=22x=e2xln2
u=2xln2, du=2ln2dx,
y=int eu/2ln2 du =
22x/2ln2 + C

conversely;
for the answer y=1/2e2x+2x+C;
y'=22x=e2xln2 = e2x+eln2 this was my mistake.
 
You have to use the logarithm rules carefully. One says that a^(m + n) = a^m * a^n, not a^(mn) = a^m + a^n.

thanks,
I was typing my reply while you were noting my error.
I did try to say amn = am + an
 
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