Logs & e

[Blade]

Need a kickstart with this one:
f(t)=t^(3/2)log(of 2)Sqrt(t+1)

Integral of (3-x)7^[(3-x)^2] dx

7^[(3-x)2] = e^[(3-x)2ln 7]
u=(3-x)^2
du/dx = -2(3-x)
(3-x)dx = -1/2 du
not even sure what so far is right..

[HallsofIvy]

I'm not sure what the question is!

"f(t)=t^(3/2)log(of 2)Sqrt(t+1)"

Okay, what's the question???


"Integral of (3-x)7^[(3-x)^2] dx

7^[(3-x)2] = e^[(3-x)2ln 7]
u=(3-x)^2
du/dx = -2(3-x)
(3-x)dx = -1/2 du
not even sure what so far is right.."

Seeing the exponent (3-x)2 and (3-x) multiplying the exponential, the first thing I would try is "let u= (3-x)2". Then du= -2(3-x)dx so the integral becomes

-2 times Integral of 7udu.

If you don't know the derivative and anti-derivative of 7u, remember that 7u= eu ln(7).

[Blade]

Ah, sorry about the first one.

f(t)=t^(3/2)log(of 2)Sqrt(t+1)
I need to derive that.

[himanshu121]

I feel u want derivative , if so then hint is

take log on both sides and then differentiate

[HallsofIvy]

I don't see any reason to take the logarithm. It's looks like a pretty direct application of the product rule and chain rule.

f(t)=t3/2(log[sub]2[/sup](√(t+1))

f'= (t3/2)'log[sub]2[/sup](√(t+1))+(t2)(log[sub]2[/sup](√(t+1))'

(t3/2)'= (3/2)t1/2, of course.

To differentiate log2(x) recall that log2(x)= ln(x)/ln(2) so (log2(x))'= 1/(xln(2)).

[himanshu121]

[:))] There are many ways of doing a problem, though both are easy to use.

yes it is a direct problem involving the product rule and chain rule[6)]