Question

Solve `4^(x+1)+3t2^x=1` for `x`?

Answer 1

`x=log_2(sqrt(9t^2+16)-3t)-3`

Explanation:

Let `2^x=y`, then `y^2=(2^x)^2=2^(2x)=(2^2)^x=4^x`

Observe that `2^x>0` i.e. `y>0`

Hence `4^(x+1)+3t2^x=1` can be written as

`4xx4^x+3t2^x=1`

or `4y^2+3ty-1=0`

and `y=(-3t+-sqrt(9t^2-4*4*(-1)))/8`

= `(-3t+-sqrt(9t^2+16))/8`

As `y>0`, we cannot have `y=(-3t-sqrt(9t^2+16))/8`

and `y=(-3t+sqrt(9t^2+16))/8` i.e. `(sqrt(9t^2+16)-3t)/8`

i.e. `2^x=(sqrt(9t^2+16)-3t)/8`

and `x=log_2((sqrt(9t^2+16)-3t)/8)`

= `log_2(sqrt(9t^2+16)-3t)-3`