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Source: — Problem Solving |
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by mtg123 » Sat Sep 01, 2007 6:50 am
2^x - 2^(x-2) = 3* 2^13
(2^(x-2))*2^2 - 2^(x-2) = 3* 2^13
(2^(x-2)) (2^2-1) =3* 2^13
3* (2^(x-2) ) = 3* 2^13
==> x-2 = 13
x=15
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by moneyman » Sun Sep 02, 2007 6:27 am
Can u pls explain the second step..I am not able to figure it out
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by mtg123 » Sun Sep 02, 2007 12:12 pm
2^x - 2^(x-2) = 3* 2^13
HERE 2^x CAN BE WRITTEN AS (2^(x-2+2)) OR (2^(x-2))*(2^2 )
[(2^(x-2))*(2^2)] - [2^(x-2)] = 3* 2^13
NOW AS (2^(x-2)) IS COMMON IN BOTH TERMS, I HAVE IT OUT
[(2^(x-2))] * [(2^2)-1] =3* 2^13
[2^(x-2)] * 3 = 3* 2^13
==> x-2 = 13
X = 15

Hope I am clear now.
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Junior | Next Rank: 30 Posts
Posts: 22
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by mtg123 » Sun Sep 02, 2007 12:35 pm
2^x - 2^(x-2) = 3* 2^13
HERE 2^x CAN BE WRITTEN AS (2^(x-2+2)) OR (2^(x-2))*(2^2 )
[(2^(x-2))*(2^2)] - [2^(x-2)] = 3* 2^13
NOW AS (2^(x-2)) IS COMMON IN BOTH TERMS, I HAVE IT OUT
[(2^(x-2))] * [(2^2)-1] =3* 2^13
[2^(x-2)] * 3 = 3* 2^13
==> x-2 = 13
X = 15

Hope I am clear now.
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by moneyman » Sun Sep 23, 2007 2:24 am
Oh yes..much much clearer now...thanks
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Plugging in could help

by wizardofwashington » Sun Sep 23, 2007 9:00 am
I'm curious why you guys didn't try to use the "plugging in" approach. Given that there are only two answer choices above the value of 13, it didn't take me more than 15 seconds to solve this.
A falling tree resounds... but a forest grows in silence...
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