# What is root of equation 2^(x^2-4x)=1/8?

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### 1 Answer

You should notice that you may write ` 8` as a power of `2` , such that:

`2^(x^2 - 4x) = 1/2^3`

Using negative power property, yields:

`2^(x^2 - 4x) = 2^(-3)`

Since the bases are equal, hence, you need to equate the exponents, such that:

`x^2 - 4x = -3`

Bringing the terms to the left side, yields:

`x^2 - 4x + 3 = 0`

You need to use quadratic formula to evaluate the solutions, such that:

`x_(1,2) = (4+-sqrt(16 - 12))/2 => x_(1,2) = (4+-sqrt4)/2`

`x_(1,2) = (4+-2)/2 => x_1 = 3 ; x_2 = 1`

You need to test the valuesÂ `x_1 = 3 ; x_2 = 1` in equation, such that:

`x_1 = 3 => 2^(3^2 - 4*3) = 1/8 => 2^(9-12) = 1/8 2^(-3) = 1/8 => 1/2^3 = 1/8 => 1/8 = 1/8`

valid

`x_2 = 1 => 2^(1^2 - 4*1) = 1/8 => 2^(1-4) = 1/8 => 2^(-3) = 1/8 => 1/2^3 = 1/8 => 1/8 = 1/8` valid

**Hence, testing the values `x_1 = 3 ; x_2 = 1` in equation yields that the equation holds for both, thus, the solutions to the given equation are **`x = 1, x = 3.`