Question

Solve for `x`? if `4=(1+x)^24`

Answer 1

`-1+2^(1/12)`

Explanation:

`4=(1+x)^24`

`root(24)4=1+x`

`4^(1/24)=1+x`

`2^(2/24)=1+x`

`2^(1/12)=1+x`

`-1+2^(1/12)=x`

Answer 2

Extend to complex numbers:

If anyone has studies complex numbers

Explanation:

`4 = (1+x)^24`

`4 = (1+x)^24 e^(2kpi i )`

as `e^(2kpi i ) =1, AA k in ZZ`

`4^(1/24) = (1+x)e^(1/12 k pi i )`

`=> 2^(1/12) = e^(1/12 k pi i) + xe^(1/12 k pi i )`

`=> 2^(1/12) - e^(1/12 k pi i ) = xe^(1/12 k pi i)`

`=> x = (2^(1/12) - e^(1/12 k pi i ) ) / e^(1/12 k pi i )`

`=> k = {0,1,2,3, ... , 22 , 23 }`

Answer 3

`x=2^(1/12)-1`

Explanation:

We can take the `24`th root of both sides to get

`4^(1/24)=1+x`

Subtracting `1` from both sides gives us

`x=4^(1/24)-1`

We can now rewrite `4` as `2^2`. This gives us

`x=2^(2*1/24)-1`

which can be simplified as

`x=2^(1/12)-1`

Hope this helps!