2^(3x+1)=2^)2/x) I don't know what to do here to get both of the xs

To solve the equation 2^(3x+1) = 2^(2/x), you can start by equating the exponents on both sides of the equation. Since the bases are equal (both are 2), you can set the exponents equal to each other:

3x + 1 = 2/x

Now we can solve for 'x':

1. Multiply both sides of the equation by 'x' to get rid of the fraction:

3x * x + 1 * x = 2

3x^2 + x = 2

2. Rearrange the equation to form a quadratic equation by moving all terms to one side:

3x^2 + x - 2 = 0

3. Now, you can either factor the quadratic equation or use the quadratic formula to solve for 'x'. In this case, factoring may be a bit challenging, so we'll solve it using the quadratic formula:

The quadratic formula is given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

For the equation 3x^2 + x - 2 = 0, the values of 'a', 'b', and 'c' are:

a = 3, b = 1, c = -2

Substituting these values into the quadratic formula, we have:

x = (-(1) ± √((1)^2 - 4(3)(-2))) / (2(3))

Simplifying further:

x = (-1 ± √(1 + 24)) / 6

x = (-1 ± √25) / 6

x = (-1 ± 5) / 6

This yields two possible values for 'x':

x = (-1 + 5) / 6 = 4/6 = 2/3

x = (-1 - 5) / 6 = -6/6 = -1

So, the equation has two solutions: x = 2/3 and x = -1.