Let f(x) = mx + b where m and b are integers with m > 0. If the solution to 2^(f(x)) = 5 is x = (ln10)/(ln8), compute the sum of m and b. I've got that...
12265
Cino Hilliard
hillcino368
Nov 1, 2006 11:06 pm
... and ... m=3,b=-1 Cino...
12266
dwittman@...
derekwittman
Nov 2, 2006 1:47 am
How did you solve for m and b? ... From: Cino Hilliard <hillcino368@...> Date: Wednesday, November 1, 2006 6:06 pm Subject: [MATH for FUN] Re: Logs -...
12267
cino hilliard
hillcino368
Nov 2, 2006 4:47 am
... Giiven 2^(mx+b) = 5. Taking the log of both sides we have mx+b log(2) = log(5). Then substituting x= log(10)/log(8), we get m(log(10)/log(8)) + b =...
12268
cino hilliard
hillcino368
Nov 2, 2006 7:16 am
... OOPs, I blew the parantheses. This should be (mx+b)log(2) = log(5) Then substituting x= log(10)/log(8), we get (mlog(10)/log(8)+b)log(2) = log(5) Since...
12269
clooneman
Nov 8, 2006 10:30 am
rotflmao Come to think of it, it's been Oct 10th since I visited this group, which sort of defeats the purpose of ever getting my password back in the first...
12270
clooneman
Nov 8, 2006 10:36 am
Mmmmm.... the bit where it says 10/10/2006???...
12271
MorphemeAddict@...
lojbaner
Nov 8, 2006 12:14 pm
In a message dated 11/8/2006 4:37:15 AM Central Standard Time, ... Hmmm, I don't think so. Nice try, though. stevo [Non-text portions of this message have been...
12272
IcerX Blade
shinobisoulxxx2
Nov 8, 2006 1:59 pm
Better yet, where's the MODERATOR on this thread? There's a little thing called booting an obvious troll. But then, that's just my take on it ...
Hello dear members, I officially announce the creation of http://www.mymathforum.com I will keep participating in Math for Fun but my main efforts will now be...
12275
dwittman@...
derekwittman
Nov 12, 2006 9:57 pm
Suppose that, for some angles x and y, sin^2(x) + cos^2 (y) = (3/2)a and cos^2(x) + sin^2(y) = (1/2)a^2 Determine the possible value of a. In English: Suppose...
12276
cino hilliard
hillcino368
Nov 12, 2006 11:51 pm
... Real values a = - 4 a = 1...
12277
clooneman
Nov 13, 2006 12:08 am
You're right, I lost a bit or reason there or something. Oct/10/2006 1:26:39 PM Central Standard Time: Just adding my two cents Nov/8/2006 4:37:15 AM Central...
12278
slim_the_dude
Nov 13, 2006 12:08 am
... Adding the two equations, you get cos²x + sin²x + cos²y + sin²y = 3a/2 + a²/2 2 = 3a/2 + a²/2 4 = 3a + a² (a+4)(a-1) = 0 a = -4 or a = 1 We can rule...
... Because of the Carl Rove math? Maybe someone here has an idea of what that is. Apparantly it worked for the last 12 years. But, well, kinda failed on Nov...
12281
Cino Hilliard
hillcino368
Nov 13, 2006 5:32 am
... (3/2)a and ... + ... squared y ... Remember ... where ... why is a = -4 not a solution? sin^2(x) + cos^2(y) = 3/2(-4) = -6 cos^2(x) + sin^2(y) = 1/2(-4) =...
12282
Peter Otzen
bogaduck
Nov 13, 2006 11:25 am
... sin^2(x) and cos^2(y) will BOTH be positive, so they cannot sum to -6. It's a bit like 2.log(x) compared to log(x^2) One of the logarithm laws says they...
... They could be 21 and 28 I feel, relying on the old 3,4,5 Peter...
12288
Peter Otzen
bogaduck
Nov 14, 2006 2:24 am
... Of course if there is no need for a vertex to be on the hypotenuse, the sides could be: 22 and root(741) 23 and 2root(174) 24 and root(649) 25 and...
12289
MorphemeAddict@...
lojbaner
Nov 14, 2006 2:40 am
In a message dated 11/13/2006 8:24:23 PM Central Standard Time, ... What is an inscribed square of a right triangle? stevo [Non-text portions of this message...
12290
slim_the_dude
Nov 14, 2006 2:51 am
... I guess when the OP asked "Suppose that, for some angles x and y,", I assumed that x and y would need to be real numbers....
12291
cino hilliard
hillcino368
Nov 14, 2006 11:18 am
... That is correct! Good approach to the solution. The side of the inscribed square s is related to the sides of the right triangle x,y as s = xy/(x+y) This...
12292
cooperpuzzles
Nov 14, 2006 4:37 pm
... inscribed ... right ... are fairly ... tenuous. ... Let s be the side of the square, h the hypotenuse of the triangle, a + s and b + s the other two sides...
12293
cino hilliard
hillcino368
Nov 14, 2006 6:51 pm
... This is correct. Using my solve program below for s=4,h=20 we get -23.72158692307 -0.674491131301148 1.04207341161294 15.3540046427582 for s = 12, h= 35 we...