1. May 13, 2005

### m0286

From your guys help, THANKS I was able to understand questions better I was hoping someone would check my answers:

for : evaluate the following logarithms:
log22^log55 I got:
log22=2^1 log55=5^1
=1 =1
therefore 1^1=1 is that correct???

Another one: solve the following logarithm equation:
log25=log2(x+32)-log2x
I got:
log25=log2(x+32)-log2x
5=log2 (x+32)/x
x+32=(5)(x)
x+32-5x
4x-32=0
x-8
therefore x=8.... is that right??????

Next: Determine the deriviative of the following functions:
d) y=2x^3 e^4x
well I figured out how to do the e^4x part it would be: 4e^4x so would the derivative be 6x^2 4e^4x????

e.) determine the derivative:
y=sq root(x^3+e^-x +5) I got:...
y=(x^3 + e^-x + 5)^1/2
dy/dx= 1/2 (x^3 + e^-x + 5)^-1/2 (3x^2-1e^-x) Now Im stuck... Is what I have right soo farr and if so can someone help me to continue.

2. May 13, 2005

### whozum

Correct, but your reasoning isnt right.

$$log_22 = 1,\ not \ 2^1.$$
$$2^{log_22} = 2$$ because [itex] log_22 = 1[/tex]

$$log_25 = log_2(x+32)-log_2x$$ which simplifies to
$$log_25 = log_2\left(\frac{x+32}{x}\right)$$

I'm not sure what you did here, but what your supposed to do is exponentiate both sides with base 2.

$$2^{log_25} = 2^{log_2\left(\frac{x+32}{x}\right)}$$ and you'll see the log's cancel out and you get:

$$5 = \frac{x+32}{x}$$ which you did get.

From there use some algebra to solve for x, you did correctly.

Don't forget the product rule. You differentiated each term correctly though.
This is all correct except you differentiated [itex] \frac{1}{e^x} [/tex] Incorrectly. Choose u = e^x and apply the chain rule.

Last edited: May 13, 2005