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4^x+2^2x=2^200 Find x. 1) 50 2) 100 3) 199 4) 199/2 5) 200

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4^x+2^2x=2^200 Find x. 1) 50 2) 100 3) 199 4) 199/2 5) 200 [#permalink] New post 19 Aug 2003, 20:35
4^x+2^2x=2^200

Find x.

1) 50
2) 100
3) 199
4) 199/2
5) 200
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 [#permalink] New post 20 Aug 2003, 02:12
199/2

as the eqn boils down to 2^2x(1+1) = 2^200
which implies 2x + 1 = 200; thus x = 199/2
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 [#permalink] New post 20 Aug 2003, 07:42
please explain I didnt understand..I did fall for the first choice.

2^2x=4^x (Is this not true)
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Great problem [#permalink] New post 20 Aug 2003, 08:23
Seems like I got a lot of these a few weeks ago on the test. Thanks for the problem

Rich
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 [#permalink] New post 20 Aug 2003, 10:41
199/2 :lol:
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 [#permalink] New post 20 Aug 2003, 20:54
satgates wrote:
please explain I didnt understand..I did fall for the first choice.

2^2x=4^x (Is this not true)


4^x = (2.2)^x = 2^2x [4^3 = (2.2).(2.2).(2.2) = 2^6]

Hence,
4^x + 2^2x = 2^200 can be written as
2^2x +2^2x = 2^200 anc this can be written as,
2.(2^2x) = 2^(2x+1) = 2^200

If 2x+1 = 200 then x = 199/2.
  [#permalink] 20 Aug 2003, 20:54
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4^x+2^2x=2^200 Find x. 1) 50 2) 100 3) 199 4) 199/2 5) 200

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