solving exponentiation equation $4^t=2^t+K$

Question

Given this equation : $4^t=2^t + K$ , with $K$ being a positive integer , how can we solve it in order to find $t$ in terms of $K$?

I started by rearranging the equation but I'm stuck at this level : $2^t(2^t-1) = K$

Answer 1

$\left(2^{t}\right)^2-2^t-K=0$, therefore after the appropriate considerations of sign $$t=\log_2\frac{1+\sqrt{1+4K}}{2}=\log_2\left(1+\sqrt{1+4K}\right)-1$$

Answer 2

Answer:

$4^t =2^t+k$ $\Rightarrow $

$2^{2t}-2^{t}-k=0$ $\Rightarrow $$ (2^{t})^2 - 2^{t}-k=0$

$\triangle=1+4k>0$

So:

$2^{t_1}=\frac{1+\sqrt{1 +4k}}{2}$

And

$2^{t_2}=\frac{1-\sqrt{1 +4k}}{2}$ that is not correct because $1-\sqrt{1 +4k}<0 $ and $ 2^{t}=e^{tln(2)}>0$

Finally :

$2^{t}=\frac{1+\sqrt{1 +4k}}{2}$ $\Rightarrow $ $t=\frac{ln(\frac{1+\sqrt{1 +4k}}{2}) }{ln2} $ $\Rightarrow $ $ t =\frac {ln(1+\sqrt{1 +4k}) }{ln(2) }-1$