Question

How do you solve `2^x= (2^-x)-(3/2)`?

Answer 1

`x = -1`

Explanation:

We will be using the following:

The quadratic formula:
`ax^2 + bx + c = 0 => x = (-b +-sqrt(b^2-4ac))/(2a)`

`ln(a^b) = bln(a)`

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

Multiplying both sides by `2^x` gives

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

`=> (2^x)^2 + (3/2)2^x - 1 = 0`

This is now a quadratic equation. Applying the quadratic formula, we obtain

`2^x = (-3/2 +- sqrt(9/4 + 4))/2=(-3/2 +- 5/2)/2 = (-3+-5)/4`

We know, however, that for all real-valued `x`, `2^x > 0` and so we can discard `2^x = -2` leaving us with `2^x = 1/2`

Now, taking the natural log of both sides and applying the property of logarithms stated above,

`ln(2^x) = ln(1/2)`

`=> xln(2) = ln(1/2)`

`=> x = ln(1/2)/ln(2)`

For more complicated values, we would be done here, however as `1/2 = 2^(-1)`

`x = ln(2^(-1))/(ln(2)) = (-1ln(2))/(ln(2)) = -1`

Thus `x = -1`

(Note that observing earlier that `1/2 = 2^-1` would have allowed us to say that `x = -1` without any work with logarithms, however the above is a more generally applicable technique which works even when the values do not work out so nicely)