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# The function f is defined by f(x)=2^x-3. If f(x)=31, then the value of

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The function f is defined by f(x)=2^x-3. If f(x)=31, then the value of  [#permalink]

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Updated on: 23 Sep 2016, 11:16
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15% (low)

Question Stats:

69% (00:52) correct 31% (00:45) wrong based on 276 sessions

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The function f is defined by f(x)=2^x-3. If f(x)=31, then the value of x is between?

A) 1 and 2
B) 2 and 3
C) 3 and 4
D) 4 and 5
E) 5 and 6

Originally posted by felippemed on 22 Sep 2016, 17:25.
Last edited by felippemed on 23 Sep 2016, 11:16, edited 2 times in total.
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Re: The function f is defined by f(x)=2^x-3. If f(x)=31, then the value of  [#permalink]

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23 Sep 2016, 06:05
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Top Contributor
2
felippemed wrote:
The function f is defined by f(x)= 2^x - 3. If f(x)=31, then the value of x is between?

A) 1 and 2
B) 2 and 3
C) 3 and 4
D) 4 and 5
E) 5 and 6

Given: f(x) = 2^x - 3
So, if f(x) = 31, we can write: 2^x - 3 = 31
Add 3 to both sides to get: 2^x = 34 [solve this equation for x]

We know that 2^5 = 32
and 2^6 = 64
Since 34 lies between 32 and 64, we can conclude that x is between 5 and 6

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Re: The function f is defined by f(x)=2^x-3. If f(x)=31, then the value of  [#permalink]

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22 Sep 2016, 17:53
If f(x) = 31, then we know that f(x)= 2^x - 3 = 31.
x = sqrt(34) from the above equation

We will get a value of x which will be greater than 5, and lesser than 6. Option E is the solution
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Re: The function f is defined by f(x)=2^x-3. If f(x)=31, then the value of  [#permalink]

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22 Sep 2016, 19:06
pushpitkc wrote:
If f(x) = 31, then we know that f(x)= 2^x - 3 = 31.
x = sqrt(34) from the above equation

We will get a value of x which will be greater than 5, and lesser than 6. Option E is the solution

f(x) = $$2^x - 3$$ Not $$x^2-3$$
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Re: The function f is defined by f(x)=2^x-3. If f(x)=31, then the value of  [#permalink]

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22 Sep 2016, 19:11
3
1
$$f(x)=2^x-3$$

==>$$31= 2^x-3$$
==>$$2^x = 34$$
==> $$2^x = 17 * 2$$
==> $$2^x = (16+1) * 2$$
==> $$2^x = (2^4+2^0) * 2$$
==> $$2^x = 2^5+2^1$$

5<x<6

Ans E.
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Re: The function f is defined by f(x)=2^x-3. If f(x)=31, then the value of  [#permalink]

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01 Jan 2017, 15:35
I have a somehow silly question, but how to know that the expression in the question is:

2^x-3 (where 3 is not part of the exponent) and not x^(2-3) where 3 is part of the exponent?
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Re: The function f is defined by f(x)=2^x-3. If f(x)=31, then the value of  [#permalink]

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01 Jan 2017, 16:41
Top Contributor
Zoser wrote:
I have a somehow silly question, but how to know that the expression in the question is:

2^x-3 (where 3 is not part of the exponent) and not x^(2-3) where 3 is part of the exponent?

Good question.
It's tough to avoid ambiguity in these forums.
Brackets would certainly help.
The prompt could have been written as: f(x)=(2^x) - 3
Alternatively, if the 3 were part of the exponent, we could have written: f(x) = 2^(x-3)

On the official GMAT, you won't have to worry about that, since the exponents will be raised.

Cheers,
Brent
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Re: The function f is defined by f(x)=2^x-3. If f(x)=31, then the value of  [#permalink]

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02 Jan 2017, 06:48
GMATPrepNow wrote:
Zoser wrote:
I have a somehow silly question, but how to know that the expression in the question is:

2^x-3 (where 3 is not part of the exponent) and not x^(2-3) where 3 is part of the exponent?

Good question.
It's tough to avoid ambiguity in these forums.
Brackets would certainly help.
The prompt could have been written as: f(x)=(2^x) - 3
Alternatively, if the 3 were part of the exponent, we could have written: f(x) = 2^(x-3)

On the official GMAT, you won't have to worry about that, since the exponents will be raised.

Cheers,
Brent

Yes you are right. And sometime it led me to a very different answer.

Thanks
Mo
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Re: The function f is defined by f(x)=2^x-3. If f(x)=31, then the value of  [#permalink]

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03 Jan 2017, 10:27
felippemed wrote:
The function f is defined by f(x)=2^x-3. If f(x)=31, then the value of x is between?

A) 1 and 2
B) 2 and 3
C) 3 and 4
D) 4 and 5
E) 5 and 6

Here is how i went abt it

(2^x) - 3 = 31
(2^x) - 3 = (2^5) - 1
(2^x) - (2^2) -1 = (2^5) - 1
(2^x) = (2^5) + (2^2)

and hence x lies between 5 and 6
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Re: The function f is defined by f(x)=2^x-3. If f(x)=31, then the value of  [#permalink]

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16 Oct 2017, 22:05
0ld wrote:
$$f(x)=2^x-3$$

==>$$31= 2^x-3$$
==>$$2^x = 34$$
==> $$2^x = 17 * 2$$
==> $$2^x = (16+1) * 2$$
==> $$2^x = (2^4+2^0) * 2$$
==> $$2^x = 2^5+2^1$$

5<x<6

Ans E.

Nice solution. Alternatively:

==> $$31= 2^x-3$$
==> $$2^x = 34$$
==> $$2^x = 32 + 2$$
==> $$2^x = 2^5 + 2^1$$
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Re: The function f is defined by f(x)=2^x-3. If f(x)=31, then the value of  [#permalink]

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19 Oct 2017, 10:22
felippemed wrote:
The function f is defined by f(x)=2^x-3. If f(x)=31, then the value of x is between?

A) 1 and 2
B) 2 and 3
C) 3 and 4
D) 4 and 5
E) 5 and 6

We can create the following equation:

2^x - 3 = 31

2^x = 34

Since 2^5 = 32 and 2^6 = 64, the value of x must be between 5 and 6.

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Re: The function f is defined by f(x)=2^x-3. If f(x)=31, then the value of &nbs [#permalink] 19 Oct 2017, 10:22
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