# Wikiversity:Help desk/archives/April 2008

## solving an equation with a variable in the exponent[edit source]

hello my problem is 1.09^4x=3.4

i've tried using the "power" property to make it 4x1.09=3.4 and solving accordingly, but i get the wrong answer

the answer is supposed to be 3.0552 rounded 4 places

i am stuck and any help would be greatly appreciated

thanks in advance --12.210.5.44 20:01, 31 March 2008 (UTC)

- I believe there is a way to do it with logarithms, but I don't recall the specifics. Of course, we can always do trial and correction for this type of problem. First, let's try some integer values for X:

x 1.09^{4x}- ----------------- 1 1.41 2 1.99 3 2.81 4 3.97

- At this point we can see that 3.4 is between 2.81 and 3.97, so our answer for X is between 3 and 4. Since 3.4 is just about in the middle of those values, let's try 3.5 for X next:

x 1.09^{4x}--- ----------------- 1 1.41 2 1.99 3 2.81 3.5 3.34 4 3.97

- So, 3.34 is just a bit low, so let's try 3.6 for X:

x 1.09^{4x}--- ----------------- 1 1.41 2 1.99 3 2.81 3.5 3.34 3.6 3.46 4 3.97

- Now we're over 3.4 just as much as we were under before, so let's try 3.55 for X and add another couple decimal points of accuracy, too:

x 1.09^{4x}--- ----------------- 1 1.41 2 1.99 3 2.81 3.5 3.34 3.55 3.3998 3.6 3.46 4 3.97

- We're really close now, just slightly under, so let's try 3.551:

x 1.09^{4x}--- ----------------- 1 1.41 2 1.99 3 2.81 3.5 3.34 3.55 3.3998 3.551 3.4010 3.6 3.46 4 3.97

- The previous value was actually much closer, so let's try 3.5502 for X this time:

x 1.09^{4x}--- ----------------- 1 1.41 2 1.99 3 2.81 3.5 3.34 3.55 3.3998 3.5502 3.40006 3.551 3.4010 3.6 3.46 4 3.97

- So, this is probably close enough. Note that the answer you listed above had the zero in the wrong place. StuRat 22:49, 31 March 2008 (UTC)

- Addendum: The solution using logs is ln(3.4)/[(4)ln(1.09)] = 3.55015127. StuRat 12:44, 3 April 2008 (UTC)

You can also solve it the following way, if you don't want to try "trial and error". 1.09^(4x)=3.4 => ln(1.09^(4x))=ln(3.4) [as long as 1.09^(4x) > 0, which is true for all x in the real number system] => (4x)(ln(1.09))=ln(3.4) => 4x = (ln(3.4))/(ln(1.09)) => x = ln(3.4)/[4(ln(1.09)]

Plug the expression on the right side of the equal sign into a log friendly calculator, and it will reply with 3.55015127 /sk

## "Permission Error" when uploading a file[edit source]

--Kevin K 22:30, 31 March 2008 (UTC)How do I upload a file? I am logged in, but when I try to upload a file, I get a "Permission Error" page that says the function is restricted to autoconfirmed users.22:30, 31 March 2008 (UTC)

- Hi Kevin, this happens because some of the functions (upload, move pages,...) are available only after a certain time, e.g. 4 days, after your account was created. Your account was created, 5:36, 30 March 2008, ----Erkan Yilmaz uses the Wikiversity:Chat (try) 22:33, 31 March 2008 (UTC)

## Sage accounting software[edit source]

I need to learn sage accounting software, can anybody help me please? i also need to know any free website where i can learn SAGE software.thanx in advance --202.70.152.179 10:48, 1 April 2008 (UTC)

- They offer training at their own site here: [1]. You can also find other training with a Google search: [2]. Here is some free online training: [3]. You may want to spend some thyme looking through the sage Google search results for other free training. :-) StuRat 14:50, 1 April 2008 (UTC)

## topology 1[edit source]

Prove that a sequentially compact subset of a metric space is totally bounded --196.200.45.194 19:16, 1 April 2008 (UTC)

- Is this homework ? If so, show us what you've done so far. StuRat 19:46, 1 April 2008 (UTC)

## topology[edit source]

1. Let A be a subset of a topological space (X,T). Show that the following are equivalent i) A is compact with respect to T ii) A is compact with respect to the relative topology T_{A} on A

2. Let A be a sequentially compact subset of a metric space X. Show that A is totally bounded.

--196.200.45.194 19:25, 1 April 2008 (UTC)

- Is this homework ? If so, show us what you've done so far. StuRat 19:46, 1 April 2008 (UTC)

## Exponential growth and decay[edit source]

--12.210.5.44 04:39, 2 April 2008 (UTC)

Hey guys I'm stuck using the formula for exponential growth and decay

the problem is:

Dead Sea Scrolls Willard Libby, a nuclear chemist from the University of Chicago, developed radiocarbon dating in the 1940s. This dating method, effective on specimens up to about 40,000 years old, works best on objects like shells, charred bones, and plants that contain carbon. Libby's first great success came in 1951 when he dated the Dead Sea Scrolls. Carbon-14 has a half life of 5730 years. If Libby found 79.3% of the original carbon-14 still present, then in about what year re the scrolls made?

If you would please please give me a full detailed explanation of how to solve this problem it would be greatly appreciated.

Thanks in advance.

- First find the number of half lives using (0.5)
^{h}= 0.793

- You could do this with trial and correction. We know that less than one half life has passed since less than half of the carbon has decayed. We can try 0.5 of a half life first. Since (0.5)
^{(0.5)}= 0.7071, that's too much decay, so let's try 0.4 half lives. (0.5)^{(0.4)}= 0.7579, that's still too much decay, so let's try 0.3 half lives. (0.5)^{(0.3)}= 0.8123, so that's too little decay. Let's make a chart:

half lives portion remaining ---------- ----------------- 0.3 0.8123 0.4 0.7579 0.5 0.7071

- 0.793 is between 0.8123 and 0.7579, and closer to 0.8123, so let's choose a value closer to 0.3 than 0.4 next. I'll pick 0.33:

half lives portion remaining ---------- ----------------- 0.3 0.8123 0.33 0.7955 0.4 0.7579 0.5 0.7071

- Now let's try 0.34:

half lives portion remaining ---------- ----------------- 0.3 0.8123 0.33 0.7955 0.34 0.7901 0.4 0.7579 0.5 0.7071

- Since 0.793 is about midway between 0.7901 and 0.7955, let's try a value midway between 0.33 and 0.34:

half lives portion remaining ---------- ----------------- 0.3 0.8123 0.33 0.7955 0.335 0.7928 0.34 0.7901 0.4 0.7579 0.5 0.7071

- Let's continue to add more accuracy to the answer:

half lives portion remaining ---------- ----------------- 0.3 0.8123 0.33 0.7955 0.334 0.7933 0.3346 0.7930 0.335 0.7928 0.34 0.7901 0.4 0.7579 0.5 0.7071

- That's close enough for me, but you can continue to refine the answer further if you wish. The final step is to multiply the number of half lives times the length of each. Thus, 0.3346(5730) = 1917 years. Subtract that from the date on which the carbon dating was done (1951) to get 34 AD. In the case of carbon dating that means that's when the tree, plants, or animals from which the scrolls were made died, as they stopped absorbing carbon then. Thus, the scrolls were made some time after 34 AD, with a suitable margin of error added. See W:carbon dating for more detail including ways to calculate the answer using logarithms. Also, if you will include the work you've done using the formula you were given for exponential growth and decay, I can help find any errors in your work. StuRat 12:07, 2 April 2008 (UTC)

Actually the way you would go about solving this is to first find the rate at which it decayed.

so you set up the formula of:

.5A_{0}=A_{0}e^{k5730}

which then you would divide out A_{0} on each side of the equation leaving you with:

.5=e^{k5730}

next you would take the natural log and use the power property to bring down k5730

ln(.5) = k5730

next you would simply divide 5730 on each side to find the rate at which it decayed:

k = ln(.5)/5730

which is : -.000120968094 or -1.20968094E^{-4}

now that you have the rate k you can input that to find out the date at which the scrolls were made

you do this by inputting the value you found for k:

.793A_{0}=A_{0}e^{-0.000120968094t}

do the same things as before divide both sides by A_{0}:

.793=e^{-0.000120968094t}

take the natural log of both sides and use the power property to bring down -0.000120968094t leaving you with:

t = ln(.793)/-0.000120968094

t = 1917.299422 approximately 1917(the year it was compared to the date he found it)

so to find the actual date the scroll was created you simply subtract the date which he discovered it from the value of t(1917)

1951-1917 = 34

the scroll was made in 34 a.d.

this is the correct way to answer the problem

- Thanks for providing the log solution. I mistakenly used the current date (I've now corrected my work). Rereading the question it does appear that the carbon dating was done in 1951. The date the carbon dating was done is, of course, the date from which we should subtract the elapsed time, not the current date or the date the objects were discovered (if different). It looks like your solution can be simplified to 1951 - [(5730)ln(.793)/ln(.5)] = 34, where ln(.793)/ln(.5) is the number of half lives I also found above (.3346 half lives). StuRat 12:24, 3 April 2008 (UTC)

## wind turbines[edit source]

--117.97.150.61 03:11, 10 April 2008 (UTC)name karthik i wanted to know in detail about the wind turbines pls help me in finding the books from wikibooks give me an idea how to download the books

- To find the books I just went to Wikibooks and typed in "wind turbine" in the search box. I found the following:

- To download a book, go to each chapter, then click on File + Save As on your web browser. However, in the cases above, you really only need one chapter (the first book only has one chapter, and the second book only has one chapter that talks about wind turbines). You might also want to check out, and/or download, the Wikipedia article: W:Wind turbine. StuRat 05:03, 10 April 2008 (UTC)

## wikiversity chat[edit source]

Dear all! I wanna involve in wikiversity chat but guys the IRC and etc etc is above my comprehension.Can anyone please tell me in simple steps?Take care! Cheers --Un-predictable 11:03, 12 April 2008 (UTC)

- I don't know how to use it either, but suggest you ask at the Colloquium, which is for questions about Wikiversity: Wikiversity:Colloquium. This help desk is for general knowledge questions, like "What's the deepest point in the ocean ?". StuRat 19:24, 22 April 2008 (UTC)