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

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New post 21 Apr 2017, 06:51
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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

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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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New post 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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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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New post 21 Apr 2017, 07:23
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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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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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New post 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.
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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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New post 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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The function f is defined by f(x)=2^(x−1)−5. If f(x)=63, then the valu  [#permalink]

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New post 12 May 2017, 12:32
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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.

Answer: 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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New post 12 Sep 2018, 10:17
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Expression can be simplified to 2^(x-3)=17

2^4=16 and 2^5=32 So,4<X-3<5 which gives 8>x>7
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Re: The function f is defined by f(x)=2^(x−1)−5. If f(x)=63, then the valu &nbs [#permalink] 12 Sep 2018, 10:17
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