Two Short Questions Homework Help

[StopWatch]

Homework Statement

The first:

I was asked to evaluate (x^x)^(x^x) at f'(2). I tried to use logarithmic differentiation and ended up with a really messy left hand side of y's and y', while the right hand side was 1 + lnx + 1 + lnx. The right answer however is apparently 2^10(1 + ln2)(1 + 2ln2) and I'm not sure where I made my mistake.

The Second:

At noon, a bacteria culture has 200 bacteria. At 1 p.m., the bacteria population has grown to 800. I have to find the time where the population is 1800 assuming exponential growth.

Homework Equations

The First: The relevant equations are above.

The Second: I know m(t) = m(0)e^kt, so I have 800 = 200e^k1 but when I take the ln of both sides and solve for k (I got ln4) I end up with 1800 = 200e^ln4(t), which doesn't give me ln3/ln2 (the purported correct answer).

I'm sure I'm just missing something very simple in both of these questions, but I really appreciate it.

Thanks in advance!

 

[eumyang]

Homework Statement

The first:

I was asked to evaluate (x^x)^(x^x) at f'(2). I tried to use logarithmic differentiation and ended up with a really messy left hand side of y's and y', while the right hand side was 1 + lnx + 1 + lnx. The right answer however is apparently 2^10(1 + ln2)(1 + 2ln2) and I'm not sure where I made my mistake.

You're going to have to show your work here, and then we can see what happened. I did get the answer you stated.

The Second:

At noon, a bacteria culture has 200 bacteria. At 1 p.m., the bacteria population has grown to 800. I have to find the time where the population is 1800 assuming exponential growth.

Homework Equations

The First: The relevant equations are above.

The Second: I know m(t) = m(0)e^kt, so I have 800 = 200e^k1 but when I take the ln of both sides and solve for k (I got ln4) I end up with 1800 = 200e^ln4(t), which doesn't give me ln3/ln2 (the purported correct answer).

I'm sure I'm just missing something very simple in both of these questions, but I really appreciate it.

Thanks in advance!

On this one I didn't get the answer you stated, so maybe there was a typo somewhere?

 

[StopWatch]

For the first one I took the natural log twice basically: ln(ln(y)) = ln((x^x)ln((x^x)))

and then on the right hand side I took ln ( (x^x)(lnx^x) ) and expanded it using log laws (I should note I used log laws to bring x^x down in the first step, as you probably noticed) to ln(x^x) + ln(ln(x^x))

Then differentiating the left hand side gets pretty messy, and the right hand side is xlnx + ln(xlnx) =

(lnx + 1) + (lnx + 1)/xlnx = ans

so, 1/y 1/lny y' = ans

I have no idea though how the 2^10 comes up.

I'll post the original statement for the second one:

At noon, a bacteria culture has 200 bacteria. At 1 p.m., the bacteria population has grown to 800. Assuming exponential growth, when will the bacteria population be 1800?

 

[eumyang]

For the first one I took the natural log twice basically: ln(ln(y)) = ln((x^x)ln((x^x)))

and then on the right hand side I took ln ( (x^x)(lnx^x) ) and expanded it using log laws (I should note I used log laws to bring x^x down in the first step, as you probably noticed) to ln(x^x) + ln(ln(x^x))

Then differentiating the left hand side gets pretty messy, and the right hand side is xlnx + ln(xlnx) =

(lnx + 1) + (lnx + 1)/xlnx = ans

so, 1/y 1/lny y' = ans

You're not done yet. If I understand you correctly, since you have two expressions equaling "ans", you have
\left(\frac{1}{y}\right)\left(\frac{1}{\ln y}\right)y' = \ln x + 1 + \frac{\ln x + 1}{x\ln x}
which looks right to me. Now you have to solve for y'. Multiply both sides by y ln y, but on the right side, you're actually multiplying by
\left( x^x \right)^{x^x} \cdot x^x \ln x^x
because that's what y ln y equals. (You may want to simplify the above before multiplying.)

Once you do that, then you substitute in 2 for x, because you're asked to find f'(2).

I'll post the original statement for the second one:

At noon, a bacteria culture has 200 bacteria. At 1 p.m., the bacteria population has grown to 800. Assuming exponential growth, when will the bacteria population be 1800?

No, I'm wondering if the answer itself (ln 3 / ln 2) is a typo, assuming you got it from the back of the textbook or the solutions manual.

 

[StopWatch]

I don't think it's likely that it's a mistake only because it's the answer to a question from a past exam. Other answers are: 1:30 pm, ln5/ln3, ln5/ln2 and ln(5/2). Quick question at a lower level: Are ln5/ln2 and ln(5/2) not the same thing by log laws? Could both consequently be written ln5 - ln2?

I was wondering the same thing regarding e in some of these growth questions. I end up with something like (2500e)^k and so I'll take the ln, and have k ln(2500) because I vaguely understand that lne = 1 and so it cancels. What I don't understand fully I think is why it can't become ln2500 + lne, by log laws, and consequently ln2500 + 1. Otherwise it feels to me like I'm allowing ln(2500e) to become ln(2500) ln(e), which isn't a law to my understanding. I appreciate your help so far by the way.

 

[eumyang]

I don't think it's likely that it's a mistake only because it's the answer to a question from a past exam. Other answers are: 1:30 pm, ln5/ln3, ln5/ln2 and ln(5/2). Quick question at a lower level: Are ln5/ln2 and ln(5/2) not the same thing by log laws? Could both consequently be written ln5 - ln2?

No, no. Only
\ln \left( \frac{5}{2} \right) = \ln 5 - \ln 2

I was wondering the same thing regarding e in some of these growth questions. I end up with something like (2500e)^k and so I'll take the ln, and have k ln(2500) because I vaguely understand that lne = 1 and so it cancels. What I don't understand fully I think is why it can't become ln2500 + lne, by log laws, and consequently ln2500 + 1. Otherwise it feels to me like I'm allowing ln(2500e) to become ln(2500) ln(e), which isn't a law to my understanding. I appreciate your help so far by the way.

I don't think you're getting these right. Also what is your original expression,
(2500e)^k
or
2500e^k?
Because they are two different expressions.

If you meant the first, then when you take the log, you get
\ln (2500e)^k = k(\ln 2500 + 1)
... but if you mean the second, then when you take the log, you get
\ln 2500e^k = \ln 2500 + k

 

[StopWatch]

That's, right, I completely forgot about that rule. I should go back and look at the proofs to really engrave it in my mind. Thanks!