Logarithm Basics: Evaluating, Solving & Derivatives - Answers Checked!

[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 I am stuck... Is what I have right soo farr and if so can someone help me to continue.

 

[whozum]

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?

Correct, but your reasoning isn't right.

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

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?

Hard to read your logs, I'm interpreting it as:

$$log_25 = log_2(x+32)-log_2x$$ 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.

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?

Don't forget the product rule. You differentiated each term correctly though.

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 I am stuck... Is what I have right soo farr and if so can someone help me to continue.

This is all correct except you differentiated [itex] \frac{1}{e^x} [/tex] Incorrectly. Choose u = e^x and apply the chain rule.

 

[M98Ranger]

Great job on evaluating and solving the logarithms! Your answers are correct.

For the derivative of y=2x^3 e^4x, you are correct that the derivative of e^4x is 4e^4x. However, for the rest of the function, you need to use the product rule. The derivative would be 2(3x^2)(e^4x) + 2x^3(4e^4x) = 6x^2e^4x + 8x^3e^4x, or factoring out the common term of 2x^3e^4x, you could also write it as 2x^3e^4x(3+4x).

For the derivative of y=sq root(x^3+e^-x +5), you are also on the right track. Your derivative is correct so far, but you can simplify it further by using the chain rule. The derivative would be 1/2 (x^3 + e^-x + 5)^-1/2 (3x^2 - e^-x). Remember that the derivative of e^-x is -e^-x. So the final answer would be 1/2 (x^3 + e^-x + 5)^-1/2 (3x^2 + e^-x). Great job on understanding the concepts and applying them correctly! Keep up the good work.