Question

How do you solve `12(1-4^x)=18`?

Answer 1

`x=-1/2+((2k+1)pi i)/ln(4)` for any integer `k`

Explanation:

Divide both sides by `12` to get:

`1-4^x = 18/12 = 3/2`

Add `4^x-3/2` to both sides to get:

`-1/2 = 4^x`

For any Real value of `x`, `4^x > 0`, so it cannot equal `-1/2`.

If we had wanted `4^x = 1/2` then we could have found:

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

and hence solution `x=-1/2`.

We can make this into a set of Complex solutions for our original problem by adding an odd multiple of `(pi i)/ln(4)`

Let `x=-1/2+((2k+1)pi i)/ln(4)` for any integer `k`

Then:

`4^x = 4^(-1/2+((2k+1)pi i)/ln(4))`

`=4^(-1/2)*4^(((2k+1)pi i)/ln(4))`

`=1/2*(e^ln(4))^(((2k+1)pi i)/ln(4))`

`=1/2*e^(ln(4)*((2k+1)pi i)/ln(4))`

`=1/2*e^((2k+1)pi i)`

`=1/2*(e^(pi i))^(2k+1)`

`=1/2*(-1)^(2k+1)`

`=-1/2`