Tricky Exponents Questions

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Tricky Exponents Questions

by tonebeeze » Wed Apr 13, 2011 12:26 pm
Can someone please walk me step by step through the simplification process of this problem. I got lost along the way...lol. Thanks!
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by confused mind » Wed Apr 13, 2011 1:03 pm
2^(x+2)-3(2^(x-1))=(5/8)(2^(16-x))
4*2^x-(3/2)*2^x=(5/2^3)((2^16)/2^x)
let 2^x=m
4*m-(3/2)*m=(5*(2^13))/m
(5/2)*m=(5*(2^13))/m
m/2=(2^13)/m
m^2=2^14

replacing m=2^x
2^(2x)=2^14
2x=14
x=7

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by GMATGuruNY » Wed Apr 13, 2011 1:15 pm
We can plug in the answer choices, which represent the value of x:

Answer choice C: x=7.
2^(7+2) - 3 * 2^(7-1) = 5/8 * 2^(16-7)

2� - 3(2�) = 5/(2³) * 2�

2�(2³ - 3) = 5 * 2�

2� * 5 = 5 * 2�. Success!

The correct answer is C.
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by kevincanspain » Wed Apr 13, 2011 1:22 pm
tonebeeze wrote:Can someone please walk me step by step through the simplification process of this problem. I got lost along the way...lol. Thanks!
The key is to factor out either 2^(x+2) or 2^(x-1) from the left side. I chose the former because it will give a fractional cooefficient

(2^(x+2))(1- 3(2^-3)) = (5/8)(2^(x+2))

Since LS=RS

x+ 2 = 16 - x

x=7
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by Ian Stewart » Wed Apr 13, 2011 2:36 pm
I replied to your post about this question on another forum, so I'll just paste the link:

https://gmatclub.com/forum/tricky-expone ... 12271.html
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

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by atulmangal » Wed Apr 13, 2011 7:20 pm
GMATGuruNY wrote:We can plug in the answer choices, which represent the value of x:

Answer choice C: x=7.
2^(7+2) - 3 * 2^(7-1) = 5/8 * 2^(16-7)

2� - 3(2�) = 5/(2³) * 2�

2�(2³ - 3) = 5 * 2�

2� * 5 = 5 * 2�. Success!

The correct answer is C.
@Mitch

Isn't this method gonna be very lengthy and time taking???

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by HSPA » Wed Apr 13, 2011 7:26 pm
In LHS : take the common value ie 2^(x-1) seperately..

2^(x-1) (8-3) = 5(2^(13-x))
x-1 = 13-x

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by Strongt » Sat Apr 23, 2011 7:56 am
Ian Stewart wrote:I replied to your post about this question on another forum, so I'll just paste the link:

https://gmatclub.com/forum/tricky-expone ... 12271.html
Brilliant!!! I can confidently say I understand how its done

Thank you