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No Solution "n"

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No Solution "n"

by mygmat.2009 » Mon Jan 18, 2010 11:06 am
Can someone help me understand why n=0 is not a solution for all 5 answers?


Each of the following equations has at least one solution EXCEPT

A. -2^n = (-2)^-n
B. 2^-n = (-2)^n
C. 2^n = (-2)^-n
D. (-2)^n = -2^n
E. (-2)^-n = -2^-n

Answer:A
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by diegow77 » Mon Jan 18, 2010 9:58 pm
This is very odd, I tried to solve this using a calculator
my answers were
A) 1 = -1
B) 1 = 1
C) 1 = 1
D) 1 = -1
E) 1 = -1

Whenever you have a negative number is in a parenthesis or a positive number the result will always be 1 and if a negative number is not in a parenthesis the result will always be -1.

That's all I know but hopefully someone can provide more input than I have.
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by fibbonnaci » Wed Jan 20, 2010 12:40 am
hey, this is a simple one.

Substitute n= 0 for all options.

Note: -2 can be written ac -1*2. so -2^0 can be written as -1* 2^0, ie. -1
But (-2)^0 = 1.

Substituting n=0 gives,
A) -1 = 1 ( not a solution)
B)1=1 (solution)
C) 1=1 (solution)
D)1 = -1 (not a solution)
E)1 = -1 (not a solution)

ie. LHS is not equal to RHS for A, D and E.

Now substitute n= 1 for A,D, E.

A) -2 = -1/2 (not a solution)
D)-2 = -2 (solution)
E) -1/2 = -1/2 (solution)

So you see only A does not have a solution. Rest all have a solution.
Thus A is the answer.
Hope this helps!
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