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Tricky Exponents Question

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Tricky Exponents Question [#permalink] New post 13 Apr 2011, 12:23
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Can someone please walk me through the simplification steps on this problem. I got lost along the way. Thanks!

If 2^{x+2} - 3(2^{x-1})=5/8(2^{16-x), what is the value of x?

a. -7
b. 6
c. 7
d. 13
e. 16
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Re: Tricky Exponents Question [#permalink] New post 13 Apr 2011, 14:18
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tonebeeze wrote:
Can someone please walk me through the simplification steps on this problem. I got lost along the way. Thanks!

If 2^{x+2} - 3(2^{x-1})=5/8(2^{16-x), what is the value of x?

a. -7
b. 6
c. 7
d. 13
e. 16


There are two general principles that you can apply to any similar problem (a problem involving numbers and powers on either side of an equation):

1) get prime bases everywhere. Your goal is to get the same base on each side of the equation, in which case the exponents must be equal
2) if you are adding or subtracting two (or more) terms with the same base, factor out the *smaller* power


We can apply principle 2) on the left side of the equation in the question above, though it isn't as straightforward as in many other questions. If you were to see, for example:

7^7 + 7^5

you'd normally factor out the smaller power, which is the 7^5 term:

7^7 + 7^5 = 7^5 (7^2 + 1) = 7^5(50) = (2)(5^2)(7^5)

Notice that to get the two terms in the brackets (the 7^2 + 1), we just reduce the exponents by 5, because we factored out 7^5.

In the question above, we have an unknown in our power, but the same principle applies. If you see something like:

2^{x+2} - 2^{x-1}

we can again factor out the smaller power, which is 2^(x-1), to get

2^{x-1} (2^3 - 1)

Notice we are just subtracting x-1 from each power to work out the exponents on the two terms in the brackets.

Now, I think the way you've written the question might be confusing to some, because it appears that the 2^{16 - x} on the right side of the equation is in the denominator of a fraction. That can't be the case; then the equation cannot possibly be true. So I'm guessing the question ought to look like this:

\begin{align}
2^{x+2} - 3(2^{x-1}) &= \left(\frac{5}{8} \right) (2^{16-x}) \\
2^{x-1} (2^3 - 3) &= \left( \frac{5}{2^3} \right) (2^{16 - x}) \\
(2^{x-1})(5) &= (5)(2^{13 - x}) \\
2^{x-1} &= 2^{13 - x} \\
x - 1 &= 13 - x \\
x &= 7
\end{align}

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Re: Tricky Exponents Question [#permalink] New post 13 Apr 2011, 18:27
Going forward please format the question properly.

Right hand side expression appears as if its 5/(8* 2^(16-x)) where as its (5/8)(2^(16-x))

2^(x+2) - 3*2^(x-1) = (5/8) (2^(16-x))

=> 2^x *(5/2) = 5* 2^(13-x)

equating two powers on both sides we have

x-1 = 13-x

=> x =7

Answer is C.
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Re: Tricky Exponents Question [#permalink] New post 13 Apr 2011, 18:34
2^x(2^2 - 3*2^x/2) = 5/8*(2^(16-x))

2^x(4 * 2 - 3)/2 = 5/8 * 2^16/2^x

=> 2^2x (5/2) = 5/2^3 * 2^16

=> 2^2x = (2)^16 + 1 - 3

=> 2^2x = 2^14

=> 2x = 14

=> x = 14/2 = 7

Answer - C
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Re: Tricky Exponents Question [#permalink] New post 15 Apr 2011, 19:56
Select a random value from choices, lets say x=6
=> 2^8 - 3.2^5 = (5/8).2^10
=> 2^5(2^3 - 3) = (5).2^7
=> 2^5(5) = (5).2^7
the LHS should give a higher value and RHS lower to balance it. So we should try the next one with a slightly higher value
Answer - C
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Re: Tricky Exponents Question [#permalink] New post 16 Apr 2011, 14:32
good explanation! IanStewart. +1
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Re: Tricky Exponents Question [#permalink] New post 16 Apr 2011, 20:15
Upon simplification:
2^{x+2} * (5/8) = 2^{16-x} * (5/8)

which implies x+2 = 16-x,
x = 7.

Answer C
Re: Tricky Exponents Question   [#permalink] 16 Apr 2011, 20:15
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