jlgdr wrote:

nverma wrote:

marcusaurelius 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

IMHO A

a) –2^n = (–2)^-n

–2^n = 1/(–2)^n

–2^n * (–2)^n = 1, Keep it. Let's solve the other options..!!

b) 2^-n = (–2)^n

1/2^n = (–2)^n

1 = (–2)^n * (2^n)

For n=0, L.H.S = R.H.S

c) 2^n = (–2)^-n

2^n = 1/ (–2)^n

(2^n) * (–2)^n = 1

For n=0, L.H.S = R.H.S

d) (–2)^n = –2^n

(–2)^n / –2^n = 1

For n=1, L.H.S = R.H.S

e) (–2)^-n = –2^-n

1/ (–2)^n = 1/–2^n

For n=1, L.H.S = R.H.S

Why did you plug n=1 for the last two, wouldn't it be easier just to plug n=0 for all and see that A has no solution?

Just want to know if there was any specific reason why you did so

Thank you

Cheers

J

PS. Would be nice if we could get this question in code format!

We need to find the equation that has no solution. What we are trying to do is find at least one solution for 4 equations. The fifth one will obviously not have any solution and will be our answer.

Options (D) and (E) do not have 0 as a solution.

So you try n = 1 on (A), (D) and (E).

n = 1 is still not a solution for (A) but it is for (D) and (E).

(D) (–2)^n = –2^n

When you put n = 0, you get

(-2)^0 = -2^0

1 = -1 which doesn't hold.

So you try n = 1

(–2)^1 = -2^1

-2 = -2

n = 1 is a solution.

Same logic for (E)

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