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The function f is defined by f(x)=2^x-3. If f(x)=31, then th

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The function f is defined by f(x)=2^x-3. If f(x)=31, then th [#permalink]  24 Jul 2020, 07:10
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85% (00:29) correct 14% (00:42) wrong based on 7 sessions
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
[Reveal] Spoiler: OA

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Founder
Joined: 18 Apr 2015
Posts: 12642
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Kudos [?]: 3171 [0], given: 11677

Re: The function f is defined by f(x)=2^x-3. If f(x)=31, then th [#permalink]  24 Jul 2020, 07:11
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Re: The function f is defined by f(x)=2^x-3. If f(x)=31, then th [#permalink]  24 Jul 2020, 07:35
Expert's post
Carcass 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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Intern
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Re: The function f is defined by f(x)=2^x-3. If f(x)=31, then th [#permalink]  27 Jul 2020, 01:45
$$31 = 2^x - 3$$
$$2^x = 34$$
$$x = \log_2 (34)$$

Don't worry, no knowledge of logarithms is needed. The key is to see that $$2^5 = 32$$ and $$2^6 = 64$$. Hence the answer is between 2 and 5.
Re: The function f is defined by f(x)=2^x-3. If f(x)=31, then th   [#permalink] 27 Jul 2020, 01:45
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