# Math Help - please tell me if these answers are correct

hi.

1- log(x+8) -logx=log(8+x)

2- log square root of X^2-4
_________________
(x-1) ^3

all this fraction goes into parenthesis.

my answer is inx+1/2in(x^2-4) -1/3 in (x-1)

3= e^3x+2=1
my answer is x = -2/3

4 - a business that develops a product in high demand calculates its revenue by the function r(t) = Ro2^t where Ro represents the initial revenue and R(t) represents the revenue after t years, how much revenue is being generated after 3 years if the company initial revenue is 3 millions? here my answer is 24.000000

thank you.

2. Originally Posted by jhonwashington
hi.

1- log(x+8) -logx=log(8+x)

2- log square root of X^2-4
_________________
(x-1) ^3

all this fraction goes into parenthesis.

my answer is inx+1/2in(x^2-4) -1/3 in (x-1)

3= e^3x+2=1
my answer is x = -2/3

4 - a business that develops a product in high demand calculates its revenue by the function r(t) = Ro2^t where Ro represents the initial revenue and R(t) represents the revenue after t years, how much revenue is being generated after 3 years if the company initial revenue is 3 millions? here my answer is 24.000000

thank you.
1. Is correct note that -8 would work as well, but due to the way that logarithms work, -8 is not within the domain, and therefore cannot be an answer (can't take log(-8) or log(0))

2. I don't understand the question or the answer, so I can't comment.

3. This is correct, Try to use parentheses to show order of operations, because e^3x+2=1 looks like $e^{3x}+2=1$ which has no solutions, but from your answer I figured it must be $e^{3x+2}=1$ in which case your answer is correct.

4. This is correct, though I haven't ever seen it written as 24.000000 (this looks to me like the integer 24, usually I see commas used to show places within long numbers such as 24,000,000)

3. Originally Posted by jhonwashington
2- log square root of X^2-4
_________________
(x-1) ^3

all this fraction goes into parenthesis.

my answer is inx+1/2in(x^2-4) -1/3 in (x-1)
ok, i'm going to guess as to what you mean here. i suppose you want to expand $\ln \left( \frac {\sqrt{x^2 - 4}}{(x - 1)^3} \right)$ ?

if that's the case, i think your answer is wrong...hard to undertsand that as well. (by the way, the "l" in "ln" is a (lower case) L not an I)

$\ln \left( \frac {\sqrt{x^2 - 4}}{(x - 1)^3} \right) = \ln \sqrt{x^2 - 4} - \ln (x - 1)^3$

$= \ln \left[ (x + 2)(x - 2) \right]^{\frac 12} - 3 \ln (x - 1)$

$= \frac 12 \ln (x + 2) + \frac 12 \ln (x - 2) - 3 \ln (x - 1)$

4. Thank you angel white and jevon for your help, I really appreciate it a lot!

yes jevon, that's the exercise I meant, how you guys do to write mathematical equations using simbols as square root or square.

just a question jevon, if you were the professor and you were grading that exercise which you just did, and if I had this answer

inx+1/2 in (x+2)(x-2) - 1/3 in (x-1)
would you consider that answer totally wrong?
thank you.

5. Originally Posted by jhonwashington
Thank you angel white and jevon for your help, I really appreciate it a lot!

yes jevon, that's the exercise I meant, how you guys do to write mathematical equations using simbols as square root or square.
we use LaTex. see here

just a question jevon, if you were the professor and you were grading that exercise which you just did, and if I had this answer

inx+1/2 in (x+2)(x-2) - 1/3 in (x-1)
would you consider that answer totally wrong?
thank you.
yes. in fact, your answer kind of confuses me, i'd be interested in seeing your process. i'd probably give you a 1/10 for creativity , ok, maybe 3/10, i'm a nice guy

6. So you know, ln is not the same as log (in your question you wrote it as "log") I don't know how much you know about logarithms, but they are basically saying finding the power that the base can be taken to in order to return what you are taking the log of).

So $log(100)$ has a base of 10, and is saying 10 to what power gives me 100. In this case, the answer is 2, because $10^{2}=100$

ln is the natural logarithm, it is the same concept as log, but it uses the irrational number "e" as the base instead of the number 10 as the base. (think of e as a number that is like pi, it is constant, and continues on forever, e is roughly 2.7183) So ln(100) means what power would I take e to in order to get 100. And the answer is $ln(100)\approx 4.6$

Try to keep them separate, in your head to help prevent mistakes.

Originally Posted by Jhevon
we use LaTex. see here

yes. in fact, your answer kind of confuses me, i'd be interested in seeing your process. i'd probably give you a 1/10 for creativity , ok, maybe 3/10, i'm a nice guy
On my test before this last one, I wrote all my derivatives to the power of p instead of an apostrophe.

So $\frac{dy}{dx}=y\prime \mbox{but I wrote } y^{p}$ because I kept misinterpreting my little p.

My instructor did not appreciate it, lol.

7. Originally Posted by angel.white
So you know, ln is not the same as log (in your question you wrote it as "log") I don't know how much you know about logarithms, but they are basically saying finding the power that the base can be taken to in order to return what you are taking the log of).

So $log(100)$ has a base of 10, and is saying 10 to what power gives me 100. In this case, the answer is 2, because $10^{2}=100$
indeed you are correct here. it was habit that i followed the user and used ln rather than log. in general, we write log to mean log to the base 10, however, in advanced math, ln happens to be the only log we like to use, and so we often write log to mean ln, by default. but i guess at the posters level, log is used to mean log to the base 10, in fact. luckily in the question i did, that did not matter, since we were not evaluating anything. so changing all the "ln" to "log" should fix it.

On my test before this last one, I wrote all my derivatives to the power of p instead of an apostrophe.

So $\frac{dy}{dx}=y\prime \mbox{but I wrote } y^{p}$ because I kept misinterpreting my little p.

My instructor did not appreciate it, lol.
lol, you meant you kept misinterpreting the apostrophe, right? that's weird, using p to replace the prime

8. Originally Posted by Jhevon
indeed you are correct here. it was habit that i followed the user and used ln rather than log. in general, we write log to mean log to the base 10, however, in advanced math, ln happens to be the only log we like to use, and so we often write log to mean ln, by default. but i guess at the posters level, log is used to mean log to the base 10, in fact. luckily in the question i did, that did not matter, since we were not evaluating anything. so changing all the "ln" to "log" should fix it.
O.o It seems like that would invite confusion.
Originally Posted by Jhevon
lol, you meant you kept misinterpreting the apostrophe, right? that's weird, using p to replace the prime
Yes, lol, sorry.

OMG!!!!
I just realized something, I use software called webassign to turn in my homework, and I realized that it treats "ln" and "log" exactly the same, (took me a while to figure this out, for a while I just thought that math itself was broken :P) Because I would use Function calculator to check my answers, and it would always write "log" instead of "ln" and then my webassign would accept "log" instead of "ln" and I didn't understand what was happening lol. I actually started plugging in values for ln and log to make absolutely certain that they were not the same. And finally when I figured out that it was "messed up" (my understanding) I sent a big email to my instructor to notify him of the problem lol. I even included screenshots (I'll attach them so you can laugh with me). He never got back to me, but I bet that this is the reason the software behaves this way. I had assumed that the compiler or the code itself was messed up, because I found three different sites which treated log the same as ln, and I figured they must all 3 be based on the same faulty math engine, or compiled with the same faulty compiler or something.

See my screenshots that I emailed to my instructor to show him our software didn't work right:
first one shows 3 answers I tried, the two accepted as right were $log(\frac{8}{3}) \mbox{ and } ln(\frac{8}{3})$

second one shows the answer I put with "log" against the expected answer with "ln" and how it accepted this as correct

and the third one shows that log(e) is treated as 1
Note: exp(1) = $e^{1}=e$, I couldn't figure out how else to use the constant "e".

9. Originally Posted by angel.white
O.o It seems like that would invite confusion.
you get used to it. it's just like how in advanced math, if you see an angle or sine/cosine etc of something, you assume it is in radians rather than degrees. mathematicians hate degrees for some reason, thus, any angle should be assumed to be in radians even if they don't say it.

Yes, lol, sorry.
that's still weird. i suppose you were mixing up prime with the 1st power? that does not happen to me, since if something is to the first power, i don't write the 1. (i see some students doing things like that, and for some reason, it annoys the out of me. seeing things like $1x$ or $x^1$)

OMG!!!!
I just realized something, I use software called webassign to turn in my homework, and I realized that it treats "ln" and "log" exactly the same, (took me a while to figure this out, for a while I just thought that math itself was broken :P) Because I would use Function calculator to check my answers, and it would always write "log" instead of "ln" and then my webassign would accept "log" instead of "ln" and I didn't understand what was happening lol. I actually started plugging in values for ln and log to make absolutely certain that they were not the same. And finally when I figured out that it was "messed up" (my understanding) I sent a big email to my instructor to notify him of the problem lol. I even included screenshots (I'll attach them so you can laugh with me). He never got back to me, but I bet that this is the reason the software behaves this way. I had assumed that the compiler or the code itself was messed up, because I found three different sites which treated log the same as ln, and I figured they must all 3 be based on the same faulty math engine, or compiled with the same faulty compiler or something.

See my screenshots that I emailed to my instructor to show him our software didn't work right:
first one shows 3 answers I tried, the two accepted as right were $log(\frac{8}{3}) \mbox{ and } ln(\frac{8}{3})$

second one shows the answer I put with "log" against the expected answer with "ln" and how it accepted this as correct

and the third one shows that log(e) is treated as 1
Note: exp(1) = $e^{1}=e$, I couldn't figure out how else to use the constant "e".
lol, well, now you know the answer. "log" is taken to mean "ln" by default when you enter calculus and above. below that level, "log" is taken to mean "log to the base 10" (of course, in the pther sciences, log still means log to the base 10, it's just math that is different...or should i say preeminent, or transcendent)

10. Originally Posted by Jhevon
that's still weird. i suppose you were mixing up prime with the 1st power? that does not happen to me, since if something is to the first power, i don't write the 1. (i see some students doing things like that, and for some reason, it annoys the out of me. seeing things like $1x$ or $x^1$)
Lol actually I kept transcribing as $x^{2}$, I think the exponent I used most frequently was 2, so when I saw an exponent I couldn't quicly decipher, my mind turned it into a 2.

As an example, read the next line as quickly as you can:

Yuor mnid wlil see waht it epxcets to see, wihch is how you raed tihs.

11. Originally Posted by angel.white
Lol actually I kept transcribing as $x^{2}$, I think the exponent I used most frequently was 2, so when I saw an exponent I couldn't quicly decipher, my mind turned it into a 2.

As an example, read the next line as quickly as you can:

Yuor mnid wlil see waht it epxcets to see, wihch is how you raed tihs.
lol, ok, i got you