Factor out a 2^(x-2):
(2^x)-(2^(x-2)) = 3*(2^13)
2^(x-2)*(2^2-1) = 3*(2^13)
2^(x-2)* 3 = 3*(2^13)
x = 15
exponents
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g_beatthegmat
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I generally find the below method easier-
Let 2^x = y. This would change the given equation to-
=> y - y*(2^-2) = 3*(2^13) (replacing 2^x by y)
=> y - y/4 = 3*(2^13)
=> 4y - y = 3*4*(2^13)
=> 3y = 3*(2^2)*(2^13) (replacing 4 by 2^2)
=> y = 2^15 (3 cancels out on both the side, power of 2 adds up to 15)
Now as y was = 2^x, and y = 2^15, we get
2^x = 2^15.
Which is x = 15
Let 2^x = y. This would change the given equation to-
=> y - y*(2^-2) = 3*(2^13) (replacing 2^x by y)
=> y - y/4 = 3*(2^13)
=> 4y - y = 3*4*(2^13)
=> 3y = 3*(2^2)*(2^13) (replacing 4 by 2^2)
=> y = 2^15 (3 cancels out on both the side, power of 2 adds up to 15)
Now as y was = 2^x, and y = 2^15, we get
2^x = 2^15.
Which is x = 15
