master_q 0 Posted March 13, 2003 As with the official boards, starbase12, & trekonline I would also like to post my Star Trek & Physics Weekly here also. At the moment ST & Physics Weekly is directly on math. It is looking at logarithmic equations. If you don’ know how to work with them, then I would suggest (because this week’s edition is another build up of understand what they are) is to go to the official boards to look thorough them. Don’t worry my editions are not always math. It is just that this topic for Star Trek & Physics Weekly is math. Edition 43: Exponential Growth & Decay Edition 44: The # e Edition 45: Intro To Logarithmic Functions Edition 46: More Logarithmic Equations Edition 47: Properties Of Logarithms Quick Intro to logs . . . Lets say you wanted to solve for an equation like 2^x = 4 For this one you can easily arrive to the solution, but using some kind of logic how can you solve it? It actually comes down to logs The first thing that I put in edition 45 was how to convert from an exponential to a logarithmic (and back & forth) 2^x=4 log base 2 of 4 = x The first one is an exponential and the second one is a logarithmic. To get the second it says log “base” 2 . . . so the base is in fact the bass of the “x” exponent. “of 4” is what it =’s. When you think about simple algebra in general you are always told to get x by it self and that is what we are in fact doing here (just with different notation). Solving “log base 2 of 4 = x” To solve this without putting it in your calculator you have to use some logic. You have to ask yourself “what power of 2 will give me 4” and then you can simply and easily arrive to 2 because 2^2 = 2*2 = 4 Let’s say that we want to convert log base 9 of 3 = x to exponential. So it would be 9^x = 3 Common Log: log base 10 Natural Log: log base e (e is about 2.718) ********************************************************* I’ll post this week’s edition and the future editions for this topic under this. I hope that you give the math a try and try to solve them (& reply). And don’t worry the next topic will not be on math and so might be easier to look at. Master Q StarTrek_Master_Q@yahoo.com Share this post Link to post Share on other sites
master_q 0 Posted March 13, 2003 Star Trek & Physics Weekly Edition 47: Properties of Logarithms {Please Reply - Questions, Extra info, What You Think. . . } ~Math Connections “The Language of the Universe” - Major Topic Note: If you don’t know how to work this math, then I understand that because there have been a few other editions on this topic that it would be hard to jump right in at the moment. And this edition is kind of shorter then the other few on this topic so the others had more examples and things to that nature. What Do You Think? / Q’s? Do you have any interesting facts about logarithms? Math Q’s: 1) Simplify: log 6 + 2 log 2 - log 3 2) {Same as 1}: 4 log base 6 of 2 + 3 log base 6 of x - log base 6 of y 3) {Same as 1}: log base 2 of 7 + 3 log base 2 of x - log base 2 of y 4) {Same as 1}: (1/4) log base 5 of 81 - (2 log base 5 of 6 - 1/2 log base 5 of 4) 5) Evaluate: log base 8 of 16 6) Expand (“un-simplify”): log { (x^5 y^2) / (2y) } ------------- Answers: 1) Simplify: log 6 + 2 log 2 - log 3 log 6 6 log 2^2 - log 3 = log (6*2^2) – log 3 = log 8 Note: log by it self is a common log (log base 10) 2) {Same as 1}: 4 log base 6 of 2 + 3 * log base 6 of x – log base 6 of y Log base 6 of 2^4 + log base 6 of y^3 – log base 6 of y = log base 6 of (16x^3) – log base y of y = log base 6 of {(16x^3)/y} 3) {Same as 1}: log base 2 of 7 + 3 log base 2 of x - log base 2 of y log base 2 of 7 + log base 2 of x^3 – log base 2 of y = log base 2 of (7x^3) – log base 2 of y = log base 2 of { (7x^3)/y } 4) {Same as 1}: (1/4) log base 5 of 81 - (2 log base 5 of 6 - 1/2 log base 5 of 4) log base 5 of (1/6) 5) Evaluate: log base 8 of 16 4/3 6) Expand (“un-simplify”): log { (x^5 y^2) / (2y) } (5 log x + 2 log y) - (log 2 + log y) ------------- Edition 47 -- Properties of Logarithms -- In this weeks edition of Star Trek & Physics Weekly by Master Q we are going to continue to look at logarithmic and exponential equations. Now we will find out how to use a calculator with logs and learn some basic properties of logs. ~~USE OF CALCULATOR~~ To solve logs on your calculator find a button that says “log” and find one that says “ln”. When we push log it is only a common log (log base 10) and when you push ln it is only a natural log (log base e). To figure out something like log base 4 of 13 what do you do? It actually comes down to a log property: log base b of n = (log n) / (log {where log n and log b are common logs} Put that together and to solve log base 4 of 13 . . . . (log 13) / (log 4) = 1.850 . . . ~~PROPERTIES~~ log base b of (uv) = log base b of u + log base b of v log base b of (u/v) = log base b u - log base b of v log base b of (u^n) = n log base b of u To understand these properties I suggest that you write them out and then look at what has been done to expand them. Using these properties you can simplify and expand an expression. **Examples) **log base 10 of 50 = log base 10 of 5 + long base 10 of 10 = log base 5 + 1 **log base 7 of (49)^x = x log base 7 of 49 = x * 2 = 2x **4 log base 2 of 3 = log base 2 of 3^4 Master Q Share this post Link to post Share on other sites
Xeroc 0 Posted March 13, 2003 You have a typo: It is: log base b of n = (log n) / (log {where log n and log b are common logs} It should be: log base b of n = (log n ) / (log b ) {where log n and log b are common logs} Those typos can be sneaky little devils! :devil: The answers are: 1) log 8 2) (log 16x^3/y) / log 6 3) (log 7x^3/y) / log 2 4) (log 1/24) / log 5 5) log 16 / log 8 = 4/3 6) 5 log x + 2 log y - (log 2 + log y) Keep up the good work Master_Q! Share this post Link to post Share on other sites
Celtic_Swimmer 0 Posted March 14, 2003 well do i feel out of place here... -Laur Share this post Link to post Share on other sites
master_q 0 Posted March 14, 2003 The answers are: 1) log 8 2) (log 16x^3/y) / log 6 3) (log 7x^3/y) / log 2 4) (log 1/24) / log 5 5) log 16 / log 8 = 4/3 6) 5 log x + 2 log y - (log 2 + log y) Keep up the good work Master_Q! They can be simplified even more so maybe you should back track (not all of them) I’ll see everyone on Sunday (I just have a little bit of time today so this will be my only post) Master Q StarTrek_Master_Q@yahoo.com Share this post Link to post Share on other sites
Xeroc 0 Posted March 14, 2003 I forget...is "log base b of a" simpler than "log a / log b"? In that case 2,3,4 would be: 2) log base 6 of 16x^3/y 3) log base 2 of 7x^3/y 4) log base 5 of 1/24 (which is also equal to about -1.975, but you said simplify, not evaluate!) Share this post Link to post Share on other sites
master_q 0 Posted March 16, 2003 Star Trek & Physics Weekly Edition 48: Solving Exponential & Logarithmic Equations {Please Reply - Comments, Extra info, What You Think. . . } ~Math Connections “The Language of the Universe” - Major Topic "This therefore is Mathematics: she reminds you of the invisible forms of the soul; she gives life to her own discoveries; she awakens the mind and purifies the intellect; she brings light to our intrinsic ideas; she abolishes oblivion and ignorance which are ours by birth." - Proculus Diadochus What Do You Think? / Q’s? For the conclusion of this topic for ST & Physics Weekly do you have any comments on this subject/topic? Math Q’s: On all solve for x 1) 4 (2^{x-3}) = 7 2) 4 log base 6 of (x-2) = 12 3) log x + log 2x = -1 4) log 5x + log (x-1) = 2 5) 4^(3x) = 8^(x+1) Edition 48 -- Solving Exponential & Logarithmic Equations -- In this weeks edition of Star Trek & Physics Weekly by Master Q we are going to continue to look at logarithmic and exponential equations. Now we will solve exponential & logarithmic equations. This will probably be the last edition for now on this topic. However in the middle of the week I’ll probably post a Star Trek & Physics Weekly Math Test for fun. ~~PROPERTIES REVIEW~~ log base b of n = ( log n ) / ( log b ) log base b of (uv) = log base b of u + log base b of v log base b of (u/v) = log base b u - log base b of v log base b of (u^n) = n log base b of u ~~SOLVING FOR “x”~~ Now we are going to finally solve x in an equation or solve that unknown using logs. To the heart they are not that hard as long as you have been keeping up a bit on the editions before this one. Example) 2^x = 5 Re-write!!! . . . . log base 2 of 5 = x Now solve “log base 2 of 5” . . . = 2.32 x = 2.32 --So what did we do there? As you can see re-writing is very important and you have to understand how to go from an exponential form to a logarithmic form and from a logarithmic to exponential (what ever the case happens to be). If you don’t remember or don’t know how to convert, then I suggest that you look over the edition were I introduce logs and how to convert because knowing how to do that is very very important. After we converted in this case we just solved that and that gave us the value of x. Example) 10^(2x-3) + 4 = 21 10^(2x-3) = 17 {Subtract 4} log 17 = 2x - 3 {convert!!} (log 17 – 3) / 2 = x {NOTE: I just said log because log base 10 is the same as just “log”} . . . x = 2.115 Example) log base 4 of x = 3 4^3=x x = 64 ~~TROUBLE AHEAD~~ How do you solve an equation like . . . 2^(x+1) = 8^(x-4) Actually there are a few ways to solve this (as are lots of other problems). One way is to make the -base- the same for both of them so it would be like #^(. . .) = #^(. . .). After that we could solve by then looking at that “. . .” and solve those. So . . . 2^(x+1) = 8^(x-4) So we know that 8 = 2^3 right? So let’s play around with that . . . 2^(x+1) = (2^3)^(x-4) And that comes down to . . . 2^(x+1) = 2^(3x-12) Now . . . x + 1 = 3x -12 x = 6.5 This is the last edition for this topic for now and I hope that you learned something. Until Next Time I’ll See You In the Q Continuum Master Q StarTrek_Master_Q@yahoo.com Share this post Link to post Share on other sites
Xeroc 0 Posted March 17, 2003 Answers: 1) log base 2 of 7/4 + 3 = x :to: x = 3.80735 2) x = 218 3) x = .15811 4: log 5x + log (x-1) = 2 log 5x(x-1) = 2 10^2 = 5x(x-1) 100 = 5x^2 - 5x 5x^2 - 5x - 100 = 0 { using the quadratic formula: } x = 5 or x = -4 5: 4^(3x) = 8^(x+1) (2^2)^(3x) = (2^3)^(x+1) 2^(6x) = 2^(3x+1) 6x = 3x + 1 3x = 1 x = 1/3 or .333333....... Share this post Link to post Share on other sites
Celtic_Swimmer 0 Posted March 23, 2003 :devil: 2 + 2 = 4... lol ~Laur :blink: Share this post Link to post Share on other sites
master_q 0 Posted March 24, 2003 Edition 48 Answers #1: 4 (2^{x-3}) = 7 x = 3.807 #2: 4 log base 6 of (x-2) = 12 x = 218 #3: log x + log 2x = -1 x = .15811 #4: log 5x + log (x-1) = 2 x = 5 or -4 #5: 4^(3x) = 8^(x+1) x = 1 NOT 1/3 Master Q StarTrek_Master_Q@yahoo.com Share this post Link to post Share on other sites
A l t e r E g o 9 Posted March 24, 2003 2 + 2 = 4... lol ~Laur :blink: Share this post Link to post Share on other sites