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Tricky Exponents Questions

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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
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, 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.
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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

IMO 7
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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
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