manjus_mailme wrote:Hi

I got this question in a MGMAT CAT exam.If any of you can solve this question using different method ,please let me know.

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

The usual method is to substitute odd and even values of n and find what the sign and absolute value of each equn is.I find this to be tedious.

The answer is A.

essentially there cant be a simpler method than substituing values for n to see which value gives us a result for the above equations...

what u should notice is that in all equations there is a (-2) ^ n or (-2) ^ -n
this equation will always change signs with respect to n... if n is even, the rsult will be +ve else it will be -ve....

just keep this in mind.... and look at the equations again... i am sure u will be able to solve it in seconds...
rest i am sure u already have the solution...

neerajkumar1_1 wrote:

manjus_mailme wrote:Hi

I got this question in a MGMAT CAT exam.If any of you can solve this question using different method ,please let me know.

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

The usual method is to substitute odd and even values of n and find what the sign and absolute value of each equn is.I find this to be tedious.

The answer is A.

essentially there cant be a simpler method than substituing values for n to see which value gives us a result for the above equations...

what u should notice is that in all equations there is a (-2) ^ n or (-2) ^ -n
this equation will always change signs with respect to n... if n is even, the rsult will be +ve else it will be -ve....

just keep this in mind.... and look at the equations again... i am sure u will be able to solve it in seconds...
rest i am sure u already have the solution...

Thanks for the reply .

Another approach:

Usually such questions have a solution which has the power = 0, and when we are substituting values 0 seems like a good choice too.

When we substitute 0 we get solutions for B, C and D.

Now to eliminate between A and E, try out other numbers. 1 solves E thus we are left with A through process of elimination.

mj78ind wrote:Another approach:

Usually such questions have a solution which has the power = 0, and when we are substituting values 0 seems like a good choice too.

When we substitute 0 we get solutions for B, C and D.

Now to eliminate between A and E, try out other numbers. 1 solves E thus we are left with A through process of elimination.

Thank you for the reply.This is a quick approach