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# 15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}

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Math Expert
Joined: 02 Sep 2009
Posts: 59587

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02 Sep 2015, 22:28
00:00

Difficulty:

25% (medium)

Question Stats:

82% (01:39) correct 18% (02:39) wrong based on 77 sessions

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$$\frac{15}{{{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}}$$

(A) $$2^{7}$$
(B) $$2^{8}$$
(C) $$2^{9}$$
(D) $$15(2^{7})$$
(E) $$15 (2^{8})$$

Kudos for a correct solution.

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Joined: 21 May 2013
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02 Sep 2015, 22:58
1
Bunuel wrote:
$$\frac{15}{{{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}}$$

(A) $$2^{7}$$
(B) $$2^{8}$$
(C) $$2^{9}$$
(D) $$15(2^{7})$$
(E) $$15 (2^{8})$$

Kudos for a correct solution.

Denominator=1/2^5+2^6+2^7+2^8=1/2^3+2^2+2+1/2^8
Fraction becomes=15*2^8/15=2^8
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03 Sep 2015, 00:02
1
Simplifying the denominator of this fraction in necessary to calculate,as below

$$\frac{15}{2^-5+2^-6+2^-7+4^-4}$$

=$$\frac{15}{2^-5+2^-6+2^-7+2^-8}$$
=$$\frac{15}{2^-5(1+2^-1+2^-2+2^-3)}$$
=$$\frac{15}{2^-5(1+1/2+1/4+1/8)}$$
=$$\frac{15}{2^-5(15/8)}$$
=$$\frac{15}{2^-5(15/2^3)}$$
=$$\frac{15}{2^-8(15)}$$
=$$\frac{1}{2^-8}$$
=2^8

Manager
Joined: 29 Jul 2015
Posts: 152

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03 Sep 2015, 08:27
1
Bunuel wrote:
$$\frac{15}{{{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}}$$

(A) $$2^{7}$$
(B) $$2^{8}$$
(C) $$2^{9}$$
(D) $$15(2^{7})$$
(E) $$15 (2^{8})$$

Kudos for a correct solution.

$$\frac{15}{({2^{-5}+2^{-6}+2^{-7}+4^{-4}})}$$

= $$\frac{15}{{{2^{-5}+2^{-6}+2^{-7}+2^{-8}}}}$$

= $$\frac{15}{{2^{-5}({{1+2^{-1}+2^{-2}+2^{-3}})}$$

= $$\frac{(2^5)(15)}{(1 + 1/2 + 1/4 + 1/8)}$$

=$$\frac{(2^5)(15)}{(15/8)}$$

= $$8(2^5)$$

= $$(2^3)(2^5)$$

= $$2^8$$

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Joined: 28 Feb 2014
Posts: 289
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04 Sep 2015, 14:30
1
$$\frac{15}{{{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}}$$ =?

The denominator becomes 2^-5(15/8)

15*(8/15)*2^-5
=2^8

(B) $$2^{8}$$
Manager
Joined: 10 Aug 2015
Posts: 102

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06 Sep 2015, 07:58
1
Solution:

Denominator = $$2^{-5} + 2^{-6} + 2^{-7} + 2^{-8}$$ = $$2^{-8}[2^{3} + 2^{2} + 2 + 1]$$ = $$2^{-8}(8+4+2+1)$$ = $$2^{-8}(15)$$

Therefore, ans = $$\frac{(15)}{(2^{-8}(15))}$$ = $$2^{8}$$

Option B
Math Expert
Joined: 02 Sep 2009
Posts: 59587

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07 Sep 2015, 03:45
Bunuel wrote:
$$\frac{15}{{{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}}$$

(A) $$2^{7}$$
(B) $$2^{8}$$
(C) $$2^{9}$$
(D) $$15(2^{7})$$
(E) $$15 (2^{8})$$

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

When you face exponents and addition, you’ll almost always need to factor out common exponential terms so that you can multiply, which is the operation that lends itself to more flexibility with exponent rules. Here, you can first take $$4^{-4}$$ and express it as $$(2^{2})^{-4}$$, which equals $$2^{-8}$$, so that you have all common bases in the denominator.

Then, in order to multiply, factor out a $$2^{-5}$$ from each term to get: $$2^{-5} (1 + 2^{-1} + 2^{-2} + 2^{-3})$$. The $$2^{-5}$$ is now a multiplicative term, which you can express also as $$\frac{1}{(2^{5})}$$, allowing you to flip it to the numerator, which becomes $$2^{5}* 15$$.

Back to the denominator, you can express $$(1 + 2^{-1} + 2^{-2} + 2^{-3})$$ as $$1 + \frac{1}{2} + \frac{1}{4}+ \frac{1}{8}$$, which equals $$1 + \frac{7}{8} = \frac{15}{8}$$.

Putting the entire fraction together, the fraction is $$\frac{15(2^{5})}{(\frac{15}{8})}$$. The 8 can be expressed as $$2^{3}$$, again providing a common exponential base, making the fraction $$\frac{15(2^{5})}{\frac{15}{2^{3}}}$$. Dividing by a fraction is the same as multiplying by its reciprocal, leaving $$\frac{15(2^{5}) * 2^{3}}{15}$$. The 15s cancel, leaving $$2^{5} * 2^{3}$$ which equals $$2^{8}$$, answer choice B.
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16 Mar 2019, 19:33
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Re: 15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}   [#permalink] 16 Mar 2019, 19:33
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