If s, u, and v are positive integers and 2^s = 2^u + 2^v, which of the following must be true?
I. s= u
II. u not equal to v
III. s>v
A) None
B) I only
C) II only
D) III only
E) II and III
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sum of unknown power
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tanyajoseph
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agps
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I. if s=u then 2^v = 0, that is never possible, No
II. u not equal to v, we don't know s, so u can be equal to v for what we know, No
III. s>v s=u+v u and v (and s) are positive integers, so u+v is always > v, Yes
answer is D
II. u not equal to v, we don't know s, so u can be equal to v for what we know, No
III. s>v s=u+v u and v (and s) are positive integers, so u+v is always > v, Yes
answer is D
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Ad_Astra_Per_Aspera
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you could just plug numbers and check:
put s = 3 then we have 2^s = 8 , so in order for 2^u + 2^v be equal to 8 u and v can take the value of 2 each since 2^2 is 4 and so sum will be 8.
From the above it follows that when s = 3 , u =2 and v=2
now look at the statements ... only III holds true.
put s = 3 then we have 2^s = 8 , so in order for 2^u + 2^v be equal to 8 u and v can take the value of 2 each since 2^2 is 4 and so sum will be 8.
From the above it follows that when s = 3 , u =2 and v=2
now look at the statements ... only III holds true.
if the answer is D, then read further....tanyajoseph wrote:If s, u, and v are positive integers and 2^s = 2^u + 2^v, which of the following must be true?
I. s= u
II. u not equal to v
III. s>v
A) None
B) I only
C) II only
D) III only
E) II and III
2^s=2^u+2^v
or, (2^s)/(2^v)=(2^u)/(2^v)+1
or, 2^(s-v)= 2^(u-v)+1
or 2^(s-v)>2^(u-v)
taking log to both side,
(s-v)* log2> (u-v)* log2
or, s-v> u-v
or s>u
case I not true
case II
take s=2, u=1, v=1
CaseIII
s>v can be proven similarly as shown above.
