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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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21 Apr 2017, 07:51
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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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08 May 2017, 23:53
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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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21 Apr 2017, 08:23
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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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23 Apr 2017, 18:05
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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23 Apr 2017, 18:11
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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12 May 2017, 13:32
4
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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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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12 Sep 2018, 11:17
1
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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The function f is defined by f(x)=2^(x−1)−5. If f(x)=63, then the valu  [#permalink]

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17 Oct 2019, 07:00
Can someone please tell me what i am doing wrong?

63= 2^(x-1)-5
<=>68=2^(x-1)
<=>2^2+2^6=2^(x-1)
<=>2+6=x-1
<=>x=9
The function f is defined by f(x)=2^(x−1)−5. If f(x)=63, then the valu   [#permalink] 17 Oct 2019, 07:00
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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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