sum of unknown power

[tanyajoseph]

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

[erdnah]

2^s = 2^u + 2^v
s = u + v

I. no
II. no
III. yes, s = v - u, and u is a positive integer and so cannot be 0!

D

[agps]

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

[blaster]

it's not correct s=u+v . you can't just sum all these figures

can someone give an explanation to the question? i can't solve that one. for me it must be A

[Ad_Astra_Per_Aspera]

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.

[real2008]

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

if the answer is D, then read further....

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.