Question

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

Answer 1

gotta do logs and exponents :)

Answer 2

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)

Answer 3

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

Answer 4

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

Answer 5

ok thanks i think i can get it now

Answer 6

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

Answer 7

still cant solve it

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

It cannot be solved. You made a mistake.

Answer 12

i probably did :) can you point it out?

Answer 13

[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.

Answer 14

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

Answer 15

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)

Answer 16

ooohh, I see it lol

Answer 17

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

Answer 18

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

Answer 19

another way of looking at it is:

1^x 2x
---- = ---
2^x 7

Answer 20

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

Answer 21

(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)

Answer 22

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

Answer 23

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

Answer 24

lol :)

Answer 25

take log3 of each side maybe as a start?

Answer 26

x = log3(2x)

x = log3(2) + log3(x)


stuff like that

Answer 27

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

Answer 28

or find a convient "base" and see if you can manipulate it......

its always fun to try the impossible :)

Answer 29

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

Answer 30

3^3^X = 3^2X

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

Answer 31

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

Answer 32

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

Answer 33

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