What is [(2^-x) - 1]  divided by [(2^x) - 1] . Solve it.

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We need to simplify the expression (2^-x -1) / (2^x -1)

First , we will rerite 2^-x = 1/2^x

==> (2^-x) -1= (1/2^x) -1 = (1-2^x)/2^x= -(2^x-1)/2^x

Now we will substitute into the expression.

==> (2^-x)-1)/ (2^x -1) = (-2^x-1)/2^x] / (2^x-1)

Now we will reduce similar terms.

==> -1/2^x / 1 -1/2^x

==> (2^-x)-1) / (2^x-1) = -1/2^x = -2^-x

Now we will try and find the solution.

==> -1/2^x = 0

Then, there is no solution for the equation because the numerator is a constant numbers and can not be zero.

Then, there is no solution.

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We have to solve (2^-x - 1) / (2^x - 1) = 0

{ Note: It is not possible to solve (2^-x - 1) / (2^x - 1), so I have equated it to 0}

(2^-x - 1) / (2^x - 1) = 0

=> [(1/2^x) - 1]/(2^x - 1) = 0

let 2^x = y

=> [1/y - 1]/[ y - 1] = 0

=>1/y - 1 = 0

=> 1/y = 1

=> y = 1

As y = 2^x, 2^x = 1

=> x = 0

But even with x = 0, the given expression gives 0/0 which is indeterminate.

Therefore, there is no solution.

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