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Exponential Problem -Problem solving help needed

Expert replies
Source: — Problem Solving |

my ans

by g2000 » Sat Jan 20, 2007 9:55 pm
Let y = 2^x and we have

y - y/4 = 3 * (2^13)
3y / 4 = 3 * (2^13)
y = 2^13 *4
y = 2^15

plug it back to the original equ.

2^15 = 2^x
x = 15 (ans)
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Re: my ans

by docxyz » Thu Jun 19, 2008 4:21 am
g2000 wrote:Let y = 2^x and we have

y - y/4 = 3 * (2^13)
3y / 4 = 3 * (2^13)
y = 2^13 *4
y = 2^15

plug it back to the original equ.

2^15 = 2^x
x = 15 (ans)


how did u get y/4?.. kindly explain.
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Re: my ans

by Ian Stewart » Thu Jun 19, 2008 5:19 am
docxyz wrote:
g2000 wrote:Let y = 2^x and we have

y - y/4 = 3 * (2^13)
3y / 4 = 3 * (2^13)
y = 2^13 *4
y = 2^15

plug it back to the original equ.

2^15 = 2^x
x = 15 (ans)


how did u get y/4?.. kindly explain.
Notice that:
2^(x-2) = (2^x)*(2^(-2)) = (2^x)*(1/4)

by the familiar exponent rules. If y = 2^x, then 2^(x-2) will equal y/4.

Still, I find it unnecessary to introduce a new letter y here. We know that:

2^x - 2^(x-2) = 3(2^13)

If you see that you can factor out a 2^(x-2) on the left side of this equation, the rest of the algebra becomes easy enough:

(2^(x-2))*(2^2 - 1) = 3(2^13)
(2^(x-2))*3 = 3(2^13)
2^(x-2) = 2^13
x-2 = 13
x = 15
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
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by Abdulla » Thu Oct 30, 2008 10:42 pm
How did you factor the left side to get this (2^(x-2))*(2^2 - 1) = 3(2^13)
Abdulla
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