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MGMAT - Exponents with negative bases

Expert replies
Source: — Problem Solving |

by kvcpk » Wed Jul 14, 2010 9:33 am
jube wrote:Each of the following equations has at least one solution EXCEPT:

-2^n = (-2)^-n
2^-n = (-2)^n
2^n = (-2)^-n
(-2)^n = -2^n
(-2)^-n = -2^-n
put n=0
A -1=1
B 1=1
C 1=1
D 1=1
E 1=-1
So the fight is between A and E.
-2^n = (-2)^-n
for any Positive value of n, LHS is integer and RHS is fraction.
for any negative value of n, LHS is fraction and RHS is integer.
Hence no solution exists.

E
(-2)^-n = -2^-n
Both LHS and RHS will be either fractions or integers.

Hence pick A.

What is OA?
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by jube » Wed Jul 14, 2010 9:40 am
Awesome! OA is indeed A.

I was getting confused between -2^n & (-2)^n -- couldn't understand the difference.

So if I'm getting this right, -x^n will always be - (x^n) i.e. we won't consider the - sign when calculating while (-x)^n will be a figure where the - sign *will* be considered when using the exponent.

Am I right?

Thanks!
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by kvcpk » Wed Jul 14, 2010 9:47 am
jube wrote:Awesome! OA is indeed A.

I was getting confused between -2^n & (-2)^n -- couldn't understand the difference.

So if I'm getting this right, -x^n will always be - (x^n) i.e. we won't consider the - sign when calculating while (-x)^n will be a figure where the - sign *will* be considered when using the exponent.

Am I right?

Thanks!
Yes.. you got it right :)
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