Question

How do you solve `(1/3)^x = -3` ?

Answer 1

`x = -1 + ((2k+1)pi)/(ln(3)) i" "` for any integer `k`

Explanation:

For any Real value of `x` we have `(1/3)^x > 0`. So there is no Real solution.

There are Complex solutions. Consider Euler's identity:

`e^(ipi) = -1`

Hence:

`e^(i(2k+1)pi) = -1" "` for any integer `k`

Now:

`(1/3)^x = (e^(ln (1/3)))^x = e^(-xln(3))`

So if `" "x = -1 + ((2k+1)pi)/(ln(3)) i" "` then we find:

`(1/3)^x = e^(-ln(3)(-1 + ((2k+1)pi)/(ln(3)) i))`

`color(white)((1/3)^x) = e^(ln(3) + (2k+1)pi i)`

`color(white)((1/3)^x) = e^(ln(3)) * e^((2k+1)pi i)`

`color(white)((1/3)^x) = 3 * (-1)`

`color(white)((1/3)^x) = -3`