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# The function f is defined by f(x)=2^(x−1)−5. If f(x)=63, then the valu

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Joined: 02 Sep 2009
Posts: 43866
The function f is defined by f(x)=2^(x−1)−5. If f(x)=63, then the valu [#permalink]

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21 Apr 2017, 06:51
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65% (hard)

Question Stats:

52% (01:08) correct 48% (01:11) wrong based on 189 sessions

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The function f is defined by $$f(x)=2^{(x−1)}−5$$. If $$f(x)=63$$, then the value of x must be:

A. Less than 5
B. Between 5 and 6
C. Between 6 and 7
D. Between 7 and 8
E. Greater than 8
[Reveal] Spoiler: OA

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Joined: 02 Mar 2017
Posts: 271
Location: India
Concentration: Finance, Marketing
Re: The function f is defined by f(x)=2^(x−1)−5. If f(x)=63, then the valu [#permalink]

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21 Apr 2017, 07:23
f(x)=63
2^(x-1) -5 =f(x)

2^(x-1) -5 = 63

2^(x-1) = 68

x-1 should be between 6 and 7 so that 2^(x-1) will lie between 64 and 128
So x should lie between 7 and 8
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Director
Joined: 12 Nov 2016
Posts: 790
Re: The function f is defined by f(x)=2^(x−1)−5. If f(x)=63, then the valu [#permalink]

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23 Apr 2017, 17:05
Bunuel wrote:
The function f is defined by $$f(x)=2^{(x−1)}−5$$. If $$f(x)=63$$, then the value of x must be:

A. Less than 5
B. Between 5 and 6
C. Between 6 and 7
D. Between 7 and 8
E. Greater than 8

Bunuel Wait this is an algebra problem? I saw there was a separate section for functions so I'm confused as to why this problem is under algebra.
Director
Joined: 12 Nov 2016
Posts: 790
Re: The function f is defined by f(x)=2^(x−1)−5. If f(x)=63, then the valu [#permalink]

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23 Apr 2017, 17:11
Bunuel wrote:
The function f is defined by $$f(x)=2^{(x−1)}−5$$. If $$f(x)=63$$, then the value of x must be:

A. Less than 5
B. Between 5 and 6
C. Between 6 and 7
D. Between 7 and 8
E. Greater than 8

To solve this question just set the two equations equal to each other

2^(x-1)-5 = 63
2(x-1) = 68
2(6) < 68 < 2(7)
But the exponent is x less than 1 so the answer should be 1 greater than each of these values

Thus D
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Re: The function f is defined by f(x)=2^(x−1)−5. If f(x)=63, then the valu [#permalink]

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08 May 2017, 22:53
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The function f is defined by f(x)=$$2^{(x−1)}$$−5. If f(x)=63, then the value of x must be:

From above equation, $$2^{(x−1)}$$−5 = 63
$$2^{(x−1)}$$ = 63+5 = 68
$$2^x$$ / 2 = 68
$$2^x$$ = 68 x 2 = 136

Since 2^7 = 128 and 2^8 = 256... Value of x must be between 7 and 8. Answer D...

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The function f is defined by f(x)=2^(x−1)−5. If f(x)=63, then the valu [#permalink]

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12 May 2017, 12:32
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Expert's post
Bunuel wrote:
The function f is defined by $$f(x)=2^{(x−1)}−5$$. If $$f(x)=63$$, then the value of x must be:

A. Less than 5
B. Between 5 and 6
C. Between 6 and 7
D. Between 7 and 8
E. Greater than 8

We are given the function f(x) = 2^(x - 1) - 5 and that f(x) = 63; thus:

63 = 2^(x - 1) - 5

68 = (2^x)/2

136 = 2^x

Since 2^7 = 128 and 2^8 = 256, x must be between 7 and 8.

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The function f is defined by f(x)=2^(x−1)−5. If f(x)=63, then the valu   [#permalink] 12 May 2017, 12:32
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