anonymous
  • anonymous
solve for x 7(1/2)^x=2x
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
gotta do logs and exponents :)
amistre64
  • amistre64
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)
amistre64
  • amistre64
remember the log(AB) = log(A) + log(B)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
also log(N^x) = x log(N)
anonymous
  • anonymous
ok thanks i think i can get it now
anonymous
  • anonymous
Erm, unless I am missing something, that cannot be solved exactly; only through iterations.
anonymous
  • anonymous
still cant solve it
amistre64
  • amistre64
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
anonymous
  • anonymous
The answer is about 1.362, but it's unlikely you'll be able to solve it much better.
amistre64
  • amistre64
maybe change of base? thats what I was thinking....
anonymous
  • anonymous
It cannot be solved. You made a mistake.
amistre64
  • amistre64
i probably did :) can you point it out?
anonymous
  • anonymous
[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.
anonymous
  • anonymous
Either way, the next line also doesn't follow.
amistre64
  • amistre64
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)
amistre64
  • amistre64
ooohh, I see it lol
amistre64
  • amistre64
log[(1^x/2^x)/x] would that be right?
anonymous
  • anonymous
newton has the right answer but im just not seeing how to solve for it
amistre64
  • amistre64
another way of looking at it is: 1^x 2x ---- = --- 2^x 7
amistre64
  • amistre64
yeah, its the playing around with the numbers and rules that I like to practice :)
anonymous
  • anonymous
(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)
anonymous
  • anonymous
Oh, wait, of course that is right, it's almost exactly as given. Doh. But yeah, no solution, bad question
anonymous
  • anonymous
im just lost on isolating the x. so say if you had 3^X=2X how would i isolate x
amistre64
  • amistre64
lol :)
amistre64
  • amistre64
take log3 of each side maybe as a start?
amistre64
  • amistre64
x = log3(2x) x = log3(2) + log3(x) stuff like that
anonymous
  • anonymous
You cannot do it. See ' Lambert W Function ' .
amistre64
  • amistre64
or find a convient "base" and see if you can manipulate it...... its always fun to try the impossible :)
anonymous
  • anonymous
Depending on the level, you are almost certainly meant to iterate this to get nearer and nearer to the root.
amistre64
  • amistre64
3^3^X = 3^2X 3^3X = 3^2X .... is that allowed?
anonymous
  • anonymous
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
anonymous
  • anonymous
(a^b)^c = a^(bc) =/= a^(b^c), in general
anonymous
  • anonymous
Wow. So you can't use logarithms for this solution effectively? No solution.. agh. >.<

Looking for something else?

Not the answer you are looking for? Search for more explanations.