solve for x
7(1/2)^x=2x

- anonymous

solve for x
7(1/2)^x=2x

- jamiebookeater

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- amistre64

gotta do logs and exponents :)

- 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

remember the log(AB) = log(A) + log(B)

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## More answers

- amistre64

also log(N^x) = x log(N)

- anonymous

ok thanks i think i can get it now

- anonymous

Erm, unless I am missing something, that cannot be solved exactly; only through iterations.

- anonymous

still cant solve it

- 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

The answer is about 1.362, but it's unlikely you'll be able to solve it much better.

- amistre64

maybe change of base? thats what I was thinking....

- anonymous

It cannot be solved. You made a mistake.

- amistre64

i probably did :) can you point it out?

- 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

Either way, the next line also doesn't follow.

- 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

ooohh, I see it lol

- amistre64

log[(1^x/2^x)/x] would that be right?

- anonymous

newton has the right answer but im just not seeing how to solve for it

- amistre64

another way of looking at it is:
1^x 2x
---- = ---
2^x 7

- amistre64

yeah, its the playing around with the numbers and rules that I like to practice :)

- 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

Oh, wait, of course that is right, it's almost exactly as given. Doh. But yeah, no solution, bad question

- anonymous

im just lost on isolating the x. so say if you had 3^X=2X how would i isolate x

- amistre64

lol :)

- amistre64

take log3 of each side maybe as a start?

- amistre64

x = log3(2x)
x = log3(2) + log3(x)
stuff like that

- anonymous

You cannot do it. See ' Lambert W Function ' .

- amistre64

or find a convient "base" and see if you can manipulate it......
its always fun to try the impossible :)

- anonymous

Depending on the level, you are almost certainly meant to iterate this to get nearer and nearer to the root.

- amistre64

3^3^X = 3^2X
3^3X = 3^2X .... is that allowed?

- 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

(a^b)^c = a^(bc) =/= a^(b^c), in general

- anonymous

Wow. So you can't use logarithms for this solution effectively? No solution.. agh. >.<

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