I am having so much trouble with these kinds of equations. Can someone please help me
Solve for 'x':
2^2x-1 - 6.2^x-1 + 4 = 0
and...
2^2x - 5.2^x+1 + 16 = 0
For the first, do you mean $\displaystyle 2^{2x - 1} - 6 \times 2^x + 4 = 0 \,$ ? Otherwise you have $\displaystyle 2^{2x - 1}$ appearing twice .......
For the second:
$\displaystyle 2^{2x} - 5 \times 2^{x+1} + 16 = 0$
$\displaystyle \Rightarrow (2^{x})^2 - 5 \times 2 \times 2^{x} + 16 = 0$
$\displaystyle \Rightarrow (2^{x})^2 - 10 \times 2^{x} + 16 = 0$
Let $\displaystyle w = 2^x$:
$\displaystyle \Rightarrow w^2 - 10w + 16 = 0$
$\displaystyle \Rightarrow (w - 2)(w - 8) = 0$
$\displaystyle \Rightarrow w = 2, \, 8$
Therefore $\displaystyle 2 = 2^x \, $ or $\displaystyle 8 = 2^x$.
Therefore $\displaystyle 2^1 = 2^x \, $ or $\displaystyle 2^3 = 2^x$.
Therefore x = .......