Given that $y = 2^x$,\n\n(a) express $4^x$ in terms of $y$.\n\n(b) Hence, or otherwise, solve\n$$8(4^x) - 9(2^{2x}) + 1 = 0$$ - Edexcel - A-Level Maths Pure - Question 8 - 2015 - Paper 1

Substituting our expression for $4^x$ into the equation, we get:\n\n$8(y^2) - 9(2^{2x}) + 1 = 0$\n\nNext, we realize that $2^{2x} = (2^x)^2 = y^2$. Thus, we can substitute this as well:\n\n$8y^2 - 9y^2 + 1 = 0$\n\nSimplifying the equation gives us:\n\n$-y^2 + 1 = 0$\n\nRearranging yields:\n\n$y^2 = 1$\n\nTaking the square root of both sides results in:\n\n$y = \\pm 1$\n\nSince $y = 2^x$, setting $y = 1$ gives us:\n\n$2^x = 1 \\Rightarrow x = 0$\n\nSetting $y = -1$ is not possible as $2^x$ cannot be negative. Therefore, the only solution is:\n\n$x = 0$