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
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)?
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
And another on another website.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
NO! Maybe the derivative of 2^(2x) = 2^(2x) / 2Ln(2) but 2^(2x) \(\displaystyle \neq\) 2^(2x) / 2Ln(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
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?
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)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)
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.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.
Yes, y=22x/2ln2 + Cfor 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?!
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 <--- 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
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!!!!!!
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
Jomo said:Yes,(dy/dx)(22x) = 2e2x