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...
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 -...
... 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 =...
... 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...
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...
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...
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...
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...
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...
... 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...
... (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) =...
... 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...
... 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...
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...
... 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...
... 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...
... 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...
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