Whatever you share, be sure that it is relevant!
Great job thus far everyone! Post your homework queries here. Help your classmates out!
Mrs. Johnson's 20152016 BC Calculus Students ROCK!
Chandan Yashraj
11/16/2010 09:02:41 am
I am confused on tonight's homework from 5.2 on the word problems. I get the SETUP of the formula and what you plug in for the first value, but not for the e. If someone can please explain the concept of "e and ln", that would be great. I completely understand logs, but I seem to have trouble as to figuring out when to use the e and ln.
Alex masiak
11/16/2010 09:58:57 am
Ln and e both apply to natural logs. When using e in a formula you do not need to plug anything in, e is a constant. On ur calculator it is with the button for division. So just plug the equation into your calculator
Madeline Zehnal
11/16/2010 12:09:16 pm
Hey Chandan you do not need to plug anything in for e, like Alex has said. Then once you find out what variable to plug in you can just enter the entire equation into your calculator. If you ever need help just go into the ARC durring lunch. It is ALOT of help. I go almost everyday, and I get most if not all of my homework done.
Sivani Aluru
11/16/2010 01:48:52 pm
yeah Chandan! I used the calculator button and it really helped me. 2nd LN gets you to e^x, and you can just enter it in from there. Hope it helps!
Kayla Simon
11/17/2010 08:03:49 am
For section 5.2 #34, the equation is f(x)=!+3^(x4). Shouldn't this simplify to 1+3^(x+4)? When I type both of these into my calculator, they are equal but x is moved 4 to the right. Can anyone explain why it isn't 4 to the left?
Nithya Sridhar
11/17/2010 08:12:47 am
I was wondering if somebody could help me with problem 58 in section 5.2?
Lindsey Stoltz
11/17/2010 08:21:27 am
I'm a little confused on #56 from 5.2. I thought it you had the same base you could set the exponents equal to each other but it doesn't seem to work in this problem. Thanks!
Jamez Hunter
11/17/2010 08:26:29 am
Kayla  Hopefully it helps.
Jamez Hunter
11/17/2010 08:29:14 am
Lindsey  The part that might be throwing you off in 56 is the (e^4)^x
Kayla Simon
11/17/2010 08:33:05 am
Nithya, if you don't want to use logs, just break 4^x down to 2^2(x) which simplifies to 2^2x. You know 2^x=3. 2^2x would be 3^2 (because to get 2^2x from 2^x, you multiply the exponant by 2). Then, multiply 3's exponant (1) by 2, getting 3^2. This gives you 1/(3^2) which simplifies to 1/9.
Kayla Simon
11/17/2010 08:35:21 am
Ok Jamez, I now understand 34. However, what's up with 42's graph? Why is it a straight line and is the domain/range all reals without any HASYs?
Jamez Hunter
11/17/2010 09:01:29 am
I did not get just a straight line, I got a curved graph that went up and to the right with a hasy of 7. The domain is all reals but the range is y<7
Angelica durski
11/17/2010 09:11:47 am
Yeah, for number 42, since it's 73e^(2X) it's just moved up 7 but their is a still a curve. Can someone explain to me, how to do number 52. I moved the 2^X to the right and got 4^X=2^X, but I'm stuck.
Jamez Hunter
11/17/2010 09:19:34 am
4^x is equal to 2^2x. Then you have the same base and can set 2x = x
Ishta
11/17/2010 10:51:51 am
I am just clarifing, is a hole the parrt you cancel off when simplifing the formula. The cancelled part is set to equal zero and then solve and that will be your hole.
Ayo adewole
11/17/2010 11:41:30 am
Hey Angelica for #52 what I did was move over the 4^x and then you can set the two to exponents equal to eachother so you end up having x=2x then when your solving it you would move the x over and set it equal to zero and you just end up having x=0. hope that helped!
Brandon Franklin
11/17/2010 11:43:16 am
I'm having trouble with number 60 in 5.2 how would you set up the problem? And do you do it with or without logs?
Jacob Hayes
11/17/2010 12:02:54 pm
For #60 i made 5^x 1/5^x then i multiplied 5^x by each side and then the equation was 1 = (3)(5^x) i then divided by 3 and had 1/3 = 5^x From there i cubed each side and got 1/27 = 5^3x
Max Jordan
11/17/2010 12:26:47 pm
Here is what I did for question #60 in section 5.2.
Sejzelle ErastusObilo
11/17/2010 12:39:44 pm
Could someone help me with setting up number 48 on the 5.2 homework please? I'm probably overthinking it, it seems rather simple just looking at it.
Sejzelle ErastusObilo
11/17/2010 12:41:31 pm
Actually I havit. Never mind :)
Kyle Wong
11/17/2010 01:05:37 pm
To solve 60 without logs, I did:
Ayo Adewole
11/17/2010 01:18:54 pm
Thanks guys, umm 50 and 56 anyone?
Ayo Adewole
11/17/2010 01:21:06 pm
Just kidding about 56
Kyle Wong
11/17/2010 01:25:17 pm
Number 50:
Ayo Adewole
11/17/2010 01:36:45 pm
Thank you!
Jose R.
11/18/2010 01:26:26 am
Does anyone know how to get started on problems 42 and 45 in chapter 5.3?
Alex Tazic
11/18/2010 07:25:32 am
Can someone hlep me with #37 on 5.3, would it just be x isgreater than 3 or can the part in the parenthesis be negative?
Jacob Hayes
11/18/2010 07:48:21 am
Jose  for number 54 you must remember that a the argument of a log must be greater than 0. Also, the function has a verticle asymtote at x=0 because it makes the problem undefined. Thereby making there 3 possible intervals for numbers to fall under:
Meghan Irby
11/18/2010 07:50:31 am
How do you do problems like number 33?
Jacob Hayes
11/18/2010 07:51:02 am
Thats 45 not 54...
Jacob hayes
11/18/2010 07:54:39 am
Use the general form for logs such that:
The Mattinator (Matt Latham)
11/18/2010 07:59:15 am
for the ones like 33 you set it equal to x because then you can turn it into a exponent type problem then isolate x and sovle for it.
Justin Temple
11/18/2010 08:14:13 am
For number 41 from 5.2, I tried to solve it and I got a weird domain and I'm not really sure what to do so can someone help?
Jacob Hayes
11/18/2010 10:12:57 am
41) For the sake of arguement, lets say that at any time i say log for this problem it is assumed to be base 1/2
Melina Moussetis
11/18/2010 10:33:53 am
Could anyone show me how to do problems 94 and 97 in section 5.3? especially part b.
Justin Temple
11/18/2010 10:49:12 am
thanks jake, and melina for 94b you have to use logs so that the inverse would be log5x = y (where the 5 is the base) because you have to switch x and y to find the inverse. For 97b you first switch x and y, then subtract 1 to get x1=2^(y3). Then you take a log so you end up with log2(x1)=y3. Then you just add 3 to get your final answer of log2(x1)+3=y (where 2 is the base). I hope that helps
Kayla Simon
11/18/2010 12:38:18 pm
Jake and Justin, but for problem 41, the answer in the book says x does not equal 1, not x>1. Why is this?
Kelly Mathesius
11/18/2010 01:19:39 pm
x does not equal 1 is correct. Since you are taking the log (x1)^2, you can plug in a number less than 1 for x and still get a positive number. For example, if you plug in 1 for x, then you get (11)^2, which equals (2)^2 which equals 4. Since that is positive, all number less than 1 are in the domain as well as the numbers greater than one. The only exclusion in the domain is when x is 1, because you cannot take the log of 0. I hope that makes sense.
Julia DiMonte
11/18/2010 01:22:57 pm
Kayla, i think it's because theres a vasy there!
Julia DiMonte
11/18/2010 01:24:36 pm
Just kidding, Kelly is right!
Lexy Neville
11/19/2010 10:19:10 am
Can anyone show me how to do number 105 in section 5.3?? I don't understand how you can solve for h.
Lindsey Stoltz
11/21/2010 07:07:59 am
Numbers 80 and 84 in 5.3?
Alex masiak
11/21/2010 07:26:08 am
Number 100 on 5.3?
Christine Troscinski
11/21/2010 07:27:42 am
hey lexy! this is what i did. Since D=miligrams and the problem states what is the time in between injections when the miligrams is 2 I put 2 in for D n the equation. The equation then looks like 2=5e^0.4h. From this equation convert it to a natural logs problem and it then looks like ln(2/5)=0.4h and then you can figure out what h is and it equals around 2.29. Since the problem is asking for the time in hours the answer would be 2 hours and around 17 minutes.
Lindsey Stoltz
11/21/2010 07:44:06 am
Alex, ok for 100 a) I just plugged in the number they gave you (.0000001) for H. Then your equation is log(.0000001) which equals 7. So your pH is 7.
Christine Troscinski
11/21/2010 07:59:39 am
hey lindsay i know how to do #84. You can change the original problem to 6^5x+3=36. Now you need to get an equal base in order to solve so you turn 36 into 6^2. Now that you have equal bases it looks like 6^5x+3=6^2. You then make the powers equal to eachother which looks like 5x+3=2 and from there you solve for x and it equals 1/5.
Anika Gupta
11/21/2010 08:14:47 am
Can someone please help me on problem 105 in 5.3? I'm really confused on how to solve it.
maddie strick
11/21/2010 10:25:42 am
can someone show me how to do #45 and #108 from 5.3
Max Jordan
11/21/2010 11:49:18 am
For #45 in 5.3 we're finding the domain of log(subscript 5)((x+1)/x)
Ayo Adewole
11/22/2010 09:29:34 am
Can someone please do number 78
Kelly Mathesius
11/22/2010 09:55:23 am
For 78, you start with the original equation of log (base x) 1/8 = 3. You rewrite this in exponential from, so you have x^3=1/8. Next, you rewrite the 1/8 as (1/2)^3. If you substitute this in, you have x^3= (1/2)^3. Take the third root of each side and you have x=1/2. Hope that helps!
Ayo Adewole
11/22/2010 10:02:15 am
thanks!
Ayo Adewole
11/22/2010 10:28:04 am
#100 b?
Kelly Mathesius
11/22/2010 10:43:56 am
For 100 b, you take the given pH, 4.2, and simply plug it into the given equation. It should look like 4.2 = log H. Multiply each side by negative one to isolate the log: 4.2=log H. Rewrite this in exponential form, so you should have 10^(4.2) = H. Then, simply plug this into your calculator and you get H = .0000631. I’m doing this without my book and just using my spiral so sorry if there are any mistakes in the notations or anything.
Sejzelle ErastusObilo
11/22/2010 11:09:35 am
Could someone help me with number 5 and 9 please? I don't know where to start with number 5, and i got as far as ln8^(1/3) on number 9.
Kelly Mathesius
11/22/2010 11:22:15 am
For 5, you take ln (2e) and separate it into ln 2 + ln e. It was given that ln 2 = a, so you have a + ln e. ln can be rewritten as log (base e). So if you take the log (base e) e, the bases cancel and that equals 1. So for 5, the answer is a + 1.
Kelly Mathesius
11/22/2010 11:25:05 am
Sorry, i meant you get 2/5b + 1/5a, not 2/5a and 1/5b.
Sejzelle ErastusObilo
11/22/2010 11:58:35 am
Oh okay, thanks Kelly! I get it now
Shivani D
11/22/2010 02:54:44 pm
Lindsey: for 80 on 5.3, since you have lne^2x=8, you would first disregard the e, as the ln of e cancels out. Then you're left with 2x=8. By dividing both sides by 2, your answer would equal, x=4.
Bethany F.
11/26/2010 04:36:32 am
for 5.4 problems 43 & 44 how do you plug that into the calculator?
maddie strick
11/28/2010 03:01:15 am
#43
Elayna Pappas
11/28/2010 07:00:42 am
Can someone please show me how to do number 25 in section 5.5?
Christine Troscinski
11/28/2010 08:02:33 am
Could some explain to me how to do number 13 and 21 in section 5.5???
Kelly Mathesius
11/28/2010 08:30:49 am
So for 25, you take the log of each side, so you get log 1.2^x = log .5^x. Bring the exponents to the front : x log 1.2 = x log .5. Subtract (–x log .5) from each side, to get: x log 1.2 – (x log .5) = 0. Factor out the x from each log: x(log 1.2 – (log .5)) = 0. Divide each side by (log 1.2 – (log .5)): x=0.
Kelly Mathesius
11/28/2010 08:43:14 am
For 13, start by rewriting the first term as (2^x)^2. When you substitute this in for the first term, your new equation looks like this: (2^x)^2 + (2^x) – 12 = 0. From here, you treat it like a quadratic equation. Factor it to get: ((2^x) + 4)((2^x) – 3) = 0. Next, you set each factor equal to 0. For the first factor, you get 2^x = 4. You can eliminate this as a possible solution because 2^x cannot be a negative number. When you set the second factor equal to 0, you get 2^x = 3. Rewrite this in log form to get x= log (base 2) 3. Use the change of base formula to plug this into the calculator: log 3/log 2 = 1.585.
Kelly Mathesius
11/28/2010 09:11:56 am
And for 21, start by taking the log of each side: log 3 ^(12x) = log 4^x. Bring the exponents to the front: (12x) log 3 = x log 4. Multiply each side by (1/x) to eliminate the x on the right side of the equation: (1/x)(12x) log 3 = log 4. Divide each side by log 3 to isolate the terms with x in them: (1/x)(12x) = log 4/log 3. Distribute the (1/x) on the left side of the equation: (1/x) – (2x/x) = log 4/log 3. The x’s cancel in the (2x/x) term, so you get: (1/x) – 2 = log 4/log 3. Add 2 to each side: 1/x = log 4/log 3 + 2. Multiply each side by x: 1 = x(log 4/log 3 + 2). Divide each side by (log 4/log 3 + 2) to isolate x: 1/(log 4/log 3 + 2) = x. When you plug that into a calculator, you get x = .307. This is kind of confusing so if you have any questions about my work just let me know!
Muhammed Alikhan
11/28/2010 12:19:21 pm
For 29 in 5.5, could you divide by 5 first to make it <2^3x=(8/5)>? I'm not sure though, because then you don't have 2 and 8 having the same base. But I guess you could do the ln thing they did in example 6.
Kayla Simon
11/29/2010 08:41:06 am
Muhammed, you can divide by 5 because it is outside the parentheses, then you take the log of each side, move the exponent for the 2^3x side and do rewrite log(8/5) as log8log5. Then solve for x.
Alex masiak
11/29/2010 08:55:04 am
can somebody explain 14 and 22 on 5.5?
Eric Cheng
11/29/2010 10:29:52 am
To solve number 14, first:
Sohaila Mali
11/29/2010 11:49:42 am
im kind of confused on number 26? could someone please explain it. thank you
Ariel Epouhe
11/29/2010 11:52:55 am
I don't get at all how to do numbers 21 and 25 from 5.5. Could someone explain it?
Melina Moussetis
11/29/2010 01:17:25 pm
Could someone please show me how to do numbers 10, 34, 38, and 42 from section 5.5?
Sejzelle ErastusObilo
11/29/2010 01:26:23 pm
For number 21, you have to use the natural log, so with the 3^12x on the left side, you bring the 12x out front so it's (12x)ln3, and then do the same for the 4^x, so then the equation looks like (12x)ln3 = (x)ln4. Then you distribute, so it's now ln32xln3=xln4. Then you have to get the x's on one side, so it's ln3=2xln3xln4. Because you're trying to solve for x, you have to factor, so then it becomes ln3=x(ln4+2ln3). Then you divide and the result should be ln3/ln4+2ln3. I hope that makes sense :)
Ariel Epouhe
11/29/2010 01:31:22 pm
Yea that helped Sejzelle. Thanks.
Sejzelle ErastusObilo
11/29/2010 01:36:51 pm
For number 25, you also use the natural log. so 1.2^x=(0.5)^x becomes xln1.2=xln0.5. Again, because you're solving for x, bring the x terms to one side, factor, and then divide. So it becomes xln1.2+xln0.5=0. But because there's a zero on the end, whatever we divide it by, it's always going to be zero, so you can just stop there...the answers is zero.
Sejzelle ErastusObilo
11/29/2010 01:50:47 pm
Sohaila, for number 26, you have to use the natural log, so the equation becomes (1+x)ln(0.3)=(2x1)ln(1.7). Then you have to distribute on both sides, so it becomes ln(0.3)+xln(0.3)=(2x)ln(1.7)(x)ln(0.3). Then you have to get the x terms on the same side, so by subtracting the xln0.3 and adding the ln1.7, the whole thing becomes ln1.7+ln0.3=2xln1.7xln.3. Since we're trying to get x by itself, you have to factor it so you can then divide, so then it's x(2ln1.7ln0.3). Divide the ln1.7+ln0.3 by what is in the parentheses and you get x=.297
Ratuja Reddy
11/30/2010 01:30:34 am
Hey, does anyone know how to solve this:
Ratuja Reddy
11/30/2010 01:34:59 am
Sorry, it's 4^(2x)4^(x+1)5=0
Shivani D
11/30/2010 01:45:22 am
For the extra credit problem, the answer is x=(log5)/(log4)1. By using the calculator, the final value of x is .16 or 4/25.
Alex Tazic
11/30/2010 07:07:46 am
to the extra credit problem its 1.16 if you use the calculator or log5/log4 without a calculator
Shivani d.
11/30/2010 07:33:28 am
Wait sorry! my explanation failed:P
Vish (V$illy) Patel and Chris Dickens =)
11/30/2010 08:44:23 am
EXTRA CREDIT IS 1.16!!!!!!!
Vish (V$illy) Patel
11/30/2010 08:57:14 am
By the way, for number 31 on the homework, what do you do after you have log base a of (((X^2)+2x3)/((X^2)+4x12))
Muhammed Alikhan
11/30/2010 09:32:47 am
Oh I got it. Thanks Kayla
Muhammed Alikhan
11/30/2010 09:36:01 am
And yea I got (log 5/log 4) for the EC too
Kyle (the Great) Wong
11/30/2010 11:07:33 am
I know that this problem was never assigned, but could someone explain how to do #7 from 5.5?
Julia DiMonte
11/30/2010 11:44:43 am
Here's number 7 Kyle...
Alec Kestler
11/30/2010 11:50:24 am
Hey can someone help me solve number 22 from section 5.5? Its: 2^(x+1)=5^(12x)
BFrank
11/30/2010 12:32:41 pm
Whoo got the e.c. 1.16! But on the homework in 5.5, number 35 I'm not sure if I found the common base correctly but I ended up with log(base)2 (x+1)squared/x=2. Is that right?
Beatrice Koka!
11/30/2010 03:17:58 pm
I had a quick question on #47 in 5.5
Alex masiak
12/1/2010 09:48:56 am
Beatrice
Becca Neuman
12/1/2010 09:53:53 am
Hey, I know some people already asked this but as far as I can tell it hasn't been answered yet; I'm a little confused with #22 on the 5.5 homework, can anyone help?
Alex Masiak
12/1/2010 09:55:13 am
Alec
Ratuja Reddy
12/1/2010 10:15:10 am
Just a quick questoin: If there's a negative sign before a log such as log12 or anything like that, it'd be the same as log12^1, right? So the negative sign is just a 1, right?
Chinar Raul
12/1/2010 10:50:01 am
Ratuja, you're right.
Chinar Raul
12/1/2010 10:53:12 am
Just to add on to what I said before:
Alec Kestler
12/1/2010 10:56:48 am
Thanks Alex!
Cody Essling
12/1/2010 12:22:02 pm
For question 62 in the review. After you take the log of both sides in the equation do you take the x^2 and x^3 to the front of the log or just the 2 and 3?
Fernando Flores
12/1/2010 12:34:18 pm
Cody,
Christian Noblett
12/1/2010 02:15:15 pm
I am having trouble finding the answer to number 38 in 5.5 can anyone help. The problem is: log(x)/log(9)+ 3 log(x)/log(3)= 14
Shivani d
12/1/2010 03:29:37 pm
For number 38, x=81.
Julianne DiLeo
12/2/2010 07:47:09 am
So can someone please explain to
Justin Temple
12/2/2010 08:45:35 am
Juliannefirst of all, you set p equal to 50 and then divide both sides to get
Kyle Cluver
12/2/2010 09:39:32 am
Can someone please help me with number 55 on tonights review hw. I don't get how it..
Mihir Surati
12/2/2010 10:08:33 am
Can anyone explain how to do that inverse problem from the review problems we did in class?
Wesley Lai
12/2/2010 10:34:13 am
While i was reviewing for the test, i got stuck on #40 on pg 328, can anyone help?
Alex Tazic
12/2/2010 11:05:14 am
WesleyFor number forty, i looked at it that log(base 2)of x is x so it would be x to the x is 4 which i know is 2. Then i said 2 equals log(base 2)of x and then exponential equation it and you get 2 to the 2 equals x. So x equals 4
Sivani Aluru
12/2/2010 12:05:28 pm
Sivani Aluru
12/2/2010 12:14:04 pm
Kyle here is what I did for number 55:
Sivani Aluru
12/2/2010 12:20:42 pm
Mihir are you talking about the one that we did where it asks us to find the inverse of f(x) = 2^(x+3) ?
Justin Temple
12/2/2010 12:58:40 pm
could someone help me with VI letter k from the review we did in class today?
Shivani D.
12/2/2010 01:09:52 pm
Can someone please show me how to do 27 on the review?
Kyle Wong!
12/2/2010 01:54:42 pm
Shivani:
Christian Noblett
12/2/2010 02:35:41 pm
Justin VI letter k is:
Shafia murad
12/2/2010 03:55:14 pm
Can someone show me how to do 24 on the review?
Chinar Raul
12/2/2010 05:23:44 pm
Shafia, after much hemming and hawing this is what I got...
Dustin Moravick
12/2/2010 10:26:18 pm
Can someone help me with #32 on the review? Much appreciated!
Robert Riley
12/3/2010 08:44:15 am
Does anyone remember how to graph
Madeline
12/6/2010 12:06:21 am
Robert,
Madeline Zehnal
12/6/2010 12:08:14 am
Hey sorry that was me that posted before
Alex Masiak
12/6/2010 08:22:33 am
number 22 on 5.6?
Kelly Mathesius
12/6/2010 11:07:28 am
For 22, I started by setting P and t equal to something. I said P=10,000 and t=1. First, I found the effective rate of the 10,000 dollars. You find this by doing A=Prt, so A=10,700=10,000*.07*1. Next, I found the equation for the compounding quarterly part of the question. By substituting the appropriate numbers into the equation, my equation looked like: A=10,000 (1+ r/4)^(4)(1). Since both the equations equal the variable A, I set them equal to each other: 10,700=10,000 (1+ r/4)^4. I divided by 10,000: 1.07 = (1+ r/4)^4. Then, I took the 4th root of each side: 1.0171 = 1 + r/4. Subtract 1: .0171 = r/4. Multiply by 4: r = 6.82. Hope that helps!
Kelly Mathesius
12/6/2010 11:10:20 am
To get the 6.82 at the end I just multiplied the decimal answer by 100 to get the percent rate. Sorry, I like skipping steps!
maddie strick
12/6/2010 11:26:09 am
5.6 #13
Ayo Adewole
12/6/2010 11:27:21 am
Kelly, where did you get the 10,700 from?
Ayo Adewole
12/6/2010 11:29:12 am
maddie for 13 you use the present
Ayo Adewole
12/6/2010 11:45:47 am
oh never mind! i figured it out.
Christine T.
12/7/2010 07:22:18 am
for 5.6 can anyone show me how to set up the equations for number 29??
Justin Temple
12/7/2010 07:34:30 am
Christine
Justin Temple
12/7/2010 07:36:15 am
sorry, in the first equation it should be 2P=P(1+(.08/12))^12t
Lindsey Stoltz
12/7/2010 07:52:26 am
Ok, for #29 from 5.6 I understand how to set it up, but I must be making a mistake in the solving part. I keep getting t=8.28
Jamez Hunter
12/7/2010 08:11:02 am
Lindsey  you have to remember that t is years, so to calculate months you just have to multiply by 12
Jamez Hunter
12/7/2010 08:25:39 am
For 41 you have to use the equation
Anika Gupta
12/7/2010 10:02:23 am
I'm also really confused on number 29 in tonight's homework. Could someone please hlep me?
Anika Gupta
12/7/2010 10:04:07 am
*help
Shivani D.
12/7/2010 11:28:02 am
Anika: For 29, the question is basically asking for t (time). Your equation would be: 2P= P( 1+ (.08/12)^12t since the question wants to know how much time will pass until your investment is doubled (2P), which is why t is unknown. To solve for t, you first divide both sides by P:
Chandan Yashraj
12/7/2010 12:40:04 pm
on #32, how do you do the first part of the problem, where it is asking the "per annum compounded monthly"? I get the compounded continuously, for the 2nd part, but I'm completely stuck on the first part for that. please help!
Chandan Yashraj
12/7/2010 12:43:23 pm
oh, and also what is the easiest way to solve #47 in tonight's homework...I tried it but I'm pretty sure I'm doing it the long way, besides the fact that I'm making things way too complicated. help on that would be great as well.
Kelly Mathesius
12/7/2010 01:37:09 pm
For the per annum compounded monthly thing, you just use the regular present value equation. So your equation is: 175=100(1+.1/12)^12t. When you solve for t (I’m assuming you know how to solve for t), you get t=5.6. I’m doing this without my book and just my spiral, so if I messed up on something let me know.
Chinar Raul
12/7/2010 01:39:18 pm
Hey Chandan,
Chinar Raul
12/7/2010 01:40:56 pm
Ha, Kelly, I like how we post a minute away from each other.
BFrank
12/7/2010 02:25:21 pm
How would you set up number 38 in 5.6? I have it set up as 3000=P(1+.03/n?)^n(.05) trying to find P but I don't know what n is. Is John only paying once for the six months? And would that be bianually?
Kayla Simon
12/8/2010 09:13:11 am
BFrank,
Jose
12/8/2010 10:11:57 am
For number 10 in section 5.7. Are you supposed to use the uninhibited radiactive decay formula? Because its giving me negative numbers...
Lexy Neville
12/8/2010 10:47:46 am
Can someone show me how to do number 13 in section 5.7? I'm confused how you set it up.
Christine T.
12/8/2010 12:17:55 pm
hey lexy! Make sure u use the radioactive decay formula. This is how i set it up to find k: .5=e^5600k, from there you change it to a natural logs problem and then it looks like ln.5=5600k and from there you can solve for k and it equals around 0.000124. With knowing what k is the next equation looks like 0.3=e^0.000124t. You then turn this into a natural logs problem and you get ln0.3=0.000124t and then you can solve for t. I got that it equals around 9709.5 years.
Beatrice! Koka
12/8/2010 01:23:20 pm
heyy so i was just wondering for #10 in section 5.7 i got the answer of 710,802.61 and the answer key says 711,115 so i was just wondering if im doing this rightt...
Ayo Adewole
12/8/2010 01:35:27 pm
beatrice, your k should equal .0589
maddie strick
12/9/2010 09:42:36 am
im confused on how you get part c. in the homework for 5.8
Julia DiMonte
12/9/2010 12:18:14 pm
I need help with number ten also! It keeps telling me that it can't graph because of the domain! helpp?
Angelica Durski
12/10/2010 06:53:42 am
For #10 part c, you have to press stat, edit, and make sure you plugged in all the correct values. In order to graph on a clear screen press stat, then go over to cacalc and hit 9 LnReg. Next you have to tell the calculator where to put your values so you do 2nd 1 comma, then 2nd 2 comma, then hit vars go over to yvars and press 1. So it should read (LnReg L1,L2,Y1) hit enter. then you should able to go to your y= and see the formula there. Next to see just ur scatter points be sure that plot is on and hit zoom and go to stat. The graph should be a curve going down.
Kyle Wong
12/10/2010 02:08:59 pm
Julia: the domain error is most likely caused by an incorrectly defined window screen. You can fix this by hitting "zoom" and selecting ZStandard (6) or ZoomFit (0).
Ashwin Chakilum
12/11/2010 06:40:34 am
@ Beatrice! Koka
Mihir Surati
12/11/2010 07:21:26 am
For #8 from section 5.8, how did you guys plug the values into L1 and L2 because whenever i try to do it, the calculator keeps on saying, "overflow."
Christine Troscinski
12/11/2010 03:44:51 pm
I kept getting that too for #8. What u need to do is substitute the years for smaller digits. So for the first year which is 1989, you change it to 0. You do the same for the rest of the years so L1 will be numbers 010. Changing this wont change the problem and now u wont get overflow!
Elayna Pappas
12/12/2010 04:05:51 am
Can someone show me how to do number 4 in section 5.7?
Shafia murad
12/12/2010 08:01:59 am
Can anyone show me how to do 19 on 5.7 and or 13 on 5.8? 12/12/2010 10:52:21 am
Can anyone explain when to use the P=e^rt formula? I just dont understand when it use used and would appreciate some clarity
Kyle Wong
12/12/2010 11:04:45 am
Elayna: to do number 4, plug in values for A(subscript 0) (the initial amount present) and A. A is going to be half of A(subscript 0) because the time you're looking for will decrease the initial amount by half.
Kyle Wong
12/12/2010 11:16:55 am
Shafia: For 19 on 5.7, you can plug the given values into your calculator by hitting "STAT" under the "DEL" button. After hitting "STAT", press "ENTER" and enter your given x values under L1 and your given y values under L2. After entering those values, hit "STAT" again, press right, and press ExpReg (number 0). This will calculate the equation of the graph.
Melina Moussetis
12/12/2010 02:03:35 pm
I have been working on this problem for days...number 22 in section 5.6. anyone know how to do it?
Alec Kestler
12/12/2010 02:04:38 pm
I know this is kind of late, but don't judge!
Confused
12/13/2010 02:53:05 am
I don't radioactive decay. Or growth. Or interest. Or cooling.
Nithya Sridhar
12/13/2010 08:50:58 am
I am a little confused about Newton's Law of Cooling, could someone please explain it to me?
Muhammed Alikhan
12/13/2010 09:49:49 am
Okay Alec. For #11, you'd use the Anod(e^kt) formula. but you'd use the fact that (.5Anod)=Anod(e^k(1690)). Divide by Anod on each side to get .5=e^1690k. when you convert this to exponential form, you get 1690k=ln(.5). divide each side by 1690, and you get k=4.1*10^4. Now you can use the actual formula. You would have A(50)=10e^50k. plug everything in and you get 9.797g.
maddie strick
12/13/2010 09:52:20 am
can someone show me how to do #22 in section 5.7?
Ratuja Reddy
12/13/2010 10:40:35 am
Maddie,
Sivani Aluru
12/13/2010 10:51:00 am
I am confused on how to start number 13 in 5.7. Thanks in advance!
Nithya Sridhar
12/13/2010 11:08:32 am
when trying to "select the best model" for an equation (ExpReg, PwrReg, etc) the "r" doesn't show up on my calculator... is there any other way of solving those types of problems?
maddie strick
12/13/2010 11:24:53 am
sivani, if you scroll up about 25 people the answer is there
Elayna Pappas
12/13/2010 11:50:35 am
Does anyone know if we need to have all the formulas memorized for the test tomorrow or is she going to give them to us?
Nithya Sridhar
12/13/2010 11:55:55 am
Elayna  she's going to give them to us
Wesley Lai
12/13/2010 11:58:35 am
elayna
Eric Cheng
12/13/2010 12:17:25 pm
For number 13 in 5.7, I get how to solve for the constant K. My question is how do you solve for the time (the second half of the problem)? Someone posted an explanation to this problem earlier and I am confused as to where they got 0.3 in the equation 0.3=e^0.000124t
Dustin Moravick
12/13/2010 12:18:25 pm
On the chapter 5 review can someone help me with the set up for number 71. Much appreciated!
Eric Cheng
12/13/2010 12:20:26 pm
Sorry, disregard my post before this! I was looking at the wrong problem.
Kelly Mathesius
12/13/2010 12:39:50 pm
For problem 71, you use the Uninhibited Radioactive Decay formula: A(t)=Ae^kt. First, you have to find what k equals. The problem tells you the halflife of carbon14 is 5600 years. So you set k equal to 5600. Since 5600 is the HALF life, then A(t) must equal a half of A. So you set A(t) equal to .5A. So your new equation to solve for k is: .5A=Ae^5600k. The A’s cancel out through division: .5=e^5600k. Rewrite this in log form: ln .5 = 5600k. Divide by 5600 to isolate k: ln .5/5600 = k. When you solve, you get that k = .000123776. When you plug this value back into the original formula, you get: A(t) = Ae^.000123776t. Since the problem gives you that the bones had 5 percent of the original amount of carbon14, A(t) must equal .05A. So when you plug this back into the equation, you get .05A = Ae^.000123776t. Now you have to solve for t. Divide by A: .05 = e^.000123776t. Rewrite in log form: ln .05 = .000123776t. Divide by .000123776 to isolate t: ln. 05/.000123776 = t. When you solve this, you get t = 24203 years.
Dustin Moravick
12/13/2010 12:57:51 pm
Thanks! Makes alot more sense now! :)
Sejzelle ErastusObilo
12/13/2010 01:29:41 pm
Could someone show me how to do the half life problems? like number 71 on the review problems.
Sejzelle ErastusObilo
12/13/2010 01:31:13 pm
NEVER MIND, it's right above :)
Sejzelle ErastusObilo
12/13/2010 01:33:57 pm
Does anyone know where to go on part b of number 66 on the review? I got to .98=logd. Thanks.
Eric Cheng
12/13/2010 01:54:37 pm
For 66, at that point, you have to know that log d means log based 10 of d.
Kyle Cluver
12/13/2010 02:38:29 pm
Can someone help me with # 73 on the review? I don't know what to do..
Bethany
12/13/2010 02:42:48 pm
I am confused on how to find the half life in problem 4 in 5.8
Becca Neuman
12/13/2010 02:45:22 pm
Does anyone know how to set up #74 on the review, page 364? I'm a little confused on that one. Thanks!
Cody Esslilng
12/13/2010 02:48:05 pm
How do you solve question # 72 in the review. I'm using the formula for newtons law of cooling, but I'm not getting the right answer. Can somebody show how to do this problem.
Robert Riley
12/13/2010 03:09:00 pm
What is the key in the zoom menu Comments are closed.

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