jeremy.hansen Wrote:Even when reading the explanations I'm confused about how to solve this problem and I am confident I would not be able to recognize or solve in another form. Hoping for prehaps a more basic understanding of what the problem is asking, how to recognize, how to attack and perhaps additional study direction.
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
Hi,
Fisrt of all just check (D) and (E), is there any typo?
Now coming to the approach.
This is a bit confusing, but patiently you can solve this question. Basically this question is checking following fundas:
1. Properties of Indices. Specifically
(i) x^(-n) = {1/(x^n)} and
(ii) (x^a) / (x^b) = x^(a-b)
2. A negative No. raised to even Powers will give a Positive No. whereas a negative No. raised to odd powers will give a Negative No.
(-2) x (-2)x(-2) x (-2) = (-2)^4 = 16 (a Positive qty)
(-2) x (-2)x(-2) = (-2)^3 = -8 (a negative qty)
3. x^0 = 1;
I will illustrate by solving example (1)
(-2)^n = (-2)^-n ............given
From above rearranging the Right hand side
(-2)^-n = [1/{(-2)^n}]
Cross-multiplying
=> {(-2)^n} x {(-2)^n} = 1
= {(-2)^2n} =1
Right hand side can be written as (-2)^0
n=0; thus solution exists.
Similarly work out for other options also.