solve the following for x.

1)e^x^2-x=e^2

I bet you mean
1)e^(x^2-x)=e^2
x^2-x = 2
x^2 - x -2 = 0
(x-2)(x+1) = 0
x = 2 or -1

Are the rest that I did do correct?

3) Suppose that the price p (in dollars) and the demand in x(thousands) of a commodity satisfy the demand equation 6p + x + xp =94. How fast is the demand changing at times when x= 4 and p=9 and the price is rising at the rate of $2 per week.

Would I take the derivative and then plug in 4 and 9?
----------------------------------------
6 dp/dt + dx/dt +x dp/dt+ p dx/dt = 0
plug
6*2 + dx/dt + 4*2 + 9 dx/dt = 0
20 + 10 dx/dt = 0
dx/dt = -2

Is number 3 considered a related rates problem?

5) evaluate: integral of x ln 4x dx

= (x^2/2) ln 4x -x^2/4

1)evaluate: integral of x ln 4x dx

1/2 x^2 + 2x^2 Is this correct?

6) evaluate the integral from 3 to 1 of x^3 - 3x +1

1/4x^4 + 3/2 x^2 Is this the correct antiderivative?

6) evaluate the integral from 3 to 1 of x^3 - 3x +1

x^4/4
- 3 x^2 /2
+ x

thank you

Is number 3 considered a related rates problem?

I suppose so.
I should have been more careful about chain rule to show you what I did.
6p + x + xp =94.

6 dp/dx + dx/dx + x dp/dx + p dx/dx = 0

times dx/dt
6 dp/dx*dx/dt + dx/dx*dx/dt + x dp/dx*dx/dt + p dx/dx*dx/dt = 0

6 dp/dt +dx/dt +x dp/dt + p dx/dt = 0

7) A bacteria culture grows exponentially according to the equation p(t)=1000e^.1t

a) what is the original amount of bacteria and what is the growth constant
----------> 1000 because e^0 = 1, k = .1

the amount is 1000 and the constant is .1
------------> right

b) what is the differential equation satisfied by p(t)

this one I am not sure about
----------> dp/dt = .1 p
then of course
dp/p = .1 dt
ln p = .1 t + C
e^lnp = p = e^(.1t + C)
p = e^C e^(.1t) = c e^(.1t)

c) what is the population after 5 days.

Just plug in 5 to the equation.
--------> right