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gotta do logs and exponents :)
take the log of each side: log[(7)(1/2)^x] = log(2x) ; now seperate the logs into addition. log(7) + log((1/2)^x) = log(2) + log(x)
remember the log(AB) = log(A) + log(B)
also log(N^x) = x log(N)
ok thanks i think i can get it now
Erm, unless I am missing something, that cannot be solved exactly; only through iterations.
still cant solve it
log(7) + x log(1/2) = log(2) + log(x) log(7) + x log(1) - x log(2) = log(2) + log(x) [x log(1) - x log(2)] -log(x) = log(2) - log(7) x log(1) - x log(2) ------- ------- = log(2/7) log(x) log(x) log[(1^x)/x] - log[(2^x)/x] = log(2/7) log [(1^x)/(2^x)] = = log(2/7) (1^x)/(2^x) = 2/7 7(1^x) = 2(2^x) this is my stuff so far :) might be helpful
The answer is about 1.362, but it's unlikely you'll be able to solve it much better.
maybe change of base? thats what I was thinking....
It cannot be solved. You made a mistake.
i probably did :) can you point it out?
[x log(1) - x log(2)] -log(x) = log(2) - log(7) x log(1) - x log(2) ------- ------- = log(2/7) log(x) log(x) Wanna explain what you did here? I may be missing it, but rest assured there is a mistake somewhere, as what you got to can be solved (exactly), but what you started with cannot.
Either way, the next line also doesn't follow.
i was thinking along the lines of log(a) - log(b) = log(a/b) log(1^x) - log(2^x) ---------------- split the fraction? log(x)
ooohh, I see it lol
log[(1^x/2^x)/x] would that be right?
newton has the right answer but im just not seeing how to solve for it
another way of looking at it is: 1^x 2x ---- = --- 2^x 7
yeah, its the playing around with the numbers and rules that I like to practice :)
(1/2)^x = 2x/7 is right, but you can't go beyond that (in elementary functions for an exact answer; only iterations will get you a near-ish result)
Oh, wait, of course that is right, it's almost exactly as given. Doh. But yeah, no solution, bad question
im just lost on isolating the x. so say if you had 3^X=2X how would i isolate x
take log3 of each side maybe as a start?
x = log3(2x) x = log3(2) + log3(x) stuff like that
You cannot do it. See ' Lambert W Function ' .
or find a convient "base" and see if you can manipulate it...... its always fun to try the impossible :)
Depending on the level, you are almost certainly meant to iterate this to get nearer and nearer to the root.
3^3^X = 3^2X 3^3X = 3^2X .... is that allowed?
Nope (and for the record, 3^x = 2x has no real solutions at all, regardless of method). 3^(3^2) = 3^9 3^(3*2) = 3^6
(a^b)^c = a^(bc) =/= a^(b^c), in general
Wow. So you can't use logarithms for this solution effectively? No solution.. agh. >.<