$2^{2x}-4(2^x)-5=0$
is the question, if I do quadratic substitution or divide by 4 first I get the right answer of 2.32, if I do it any other way using logs I get the wrong answer. No matter if I use Log2, Ln, split it different ways, divide different ways. Always either get the answer to be negative of what it actually is or 4.32 or sometimes 0.32.
Here is one way I tried...
$2^{2x}-4(2^x)-5=0$
$2^{2x}-4(2^x)=5$
$2^{2x}-2^{2+x}=5$
$log_2(2^{2x})-log_2(2^{2+x})=log_2(5)$
$2x-(2+x)=log_2(5)$
$x-2=log_2(5)$
$x=4.32$
Clearly every time I'm doing it I'm doing something wrong but I have spent a good few hours and now I just want to know what it is.
Thank you
In your case, and after the mistake has been commented:
$$2^{2x}-4\cdot2^x-5=0\implies (2^x-5)(2^x+1)=0\stackrel{\text{since}\,2^x+1>1}\implies 2^x=5\implies x=\frac{\log5}{\log2}$$