# what is solution x, 2^x=2^(-x)?

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You need to solve exponential equation `2^x = 2^(-x).`

Use the property `2^(-x) = 1/2^x` such that:

`2^x = 1/2^x`

Subtracting `1/2^x` both sides yields:

`2^x - 1/2^x ` = 0

You need to bring the terms to a common denominator such that:

`(2^(2x) - 1)/2^x` = 0 => `2^(2x)` - 1 = 0

You need to use the product that results from a difference of squares:

`2^(2x)` - 1 = (`2^x` - 1)(`2^x ` + 1) = 0

Cancelling the first factor yields:

`2^x` - 1= 0 => `2^x` = 1

You need to remember that any number raised to 0 power yields 1 => x = 0.

`2^x` + 1 = 0 => `2^x ` = -1

This equation does not hold for any x since `2^x` > 0.

**The solution to this equation is x = 0.**