How do you solve `4^x - 2^x = 0`?

`4^x-2^x=0`

`4^x=2^x`

Apply the natural logarithm to both sides:

`ln(4^x)=ln(2^x)`

Recalling that `ln(a^b)=bln(a)`:

`xln(4)=xln(2)`

So, `x=0` is obviously a solution as it will result in `0=0.`

Beyond that, there are no solutions as `x` cancels out and we're left with

`ln(4)=ln(2)` which is not true.

`4^x-2^x=0`

now
`4=2^2=>4^x=2^(2x)`

`4^x-2^x=0=>2^(2x)-2^x=0`

factorising

`2^x(2^x-1)=0`

either `2^x=0=>`no real solns

or

`2^x-1=0`

`2^x=1=>x=0`

The other answers on the page are correct; I just wanted to display another method to solve this problem:

`4^x-2^x=0`

`4^x=2^x`

`(2^2)^x=2^x`

`2^(color(blue)(2x))=2^color(blue)x`

Since the bases are equivalent, the exponents must also be equivalent:

`color(blue)(2x)=color(blue)x`

`x=0`

Given:

`4^x-2^x = 0`

Note that `4^x = (2 * 2)^x = 2^x * 2^x`

So we have:

`0 = 4^x-2^x = 2^x(2^x-1)`

Note that `2^x = 0` has no solutions (real or complex).

So:

`2^x = 1`

This has real solution `x = 0`

How about complex solutions?

Note that `e^(2pii) = 1`, so `e^(2kpii) = 1` for any integer `k`. These are the only complex values for which `e^t = 1`.

So we find:

`e^(2kpii) = 1 = 2^x = (e^(ln 2))^x = e^(x ln 2)`

So:

`x ln 2 = 2kpii`

So:

`x = (2kpii)/ln 2" "` for any integer `k`