A-Level Edexcel Maths Pure Arithmetic Sequences & Series: Given that $y = 2^x$, (a) expre

Given that $y = 2^x$, (a) express $4^x$ in terms of $y$ - Edexcel - A-Level Maths Pure - Question 9 - 2015 - Paper 1

Question 9

Given that $y = 2^x$, (a) express $4^x$ in terms of $y$. (b) Hence, or otherwise, solve $$8(4^x) - 9(2^x) + 1 = 0$$

Worked Solution & Example Answer:Given that $y = 2^x$, (a) express $4^x$ in terms of $y$ - Edexcel - A-Level Maths Pure - Question 9 - 2015 - Paper 1

Step 1

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Answer

To express $4^x$ in terms of $y$ , we can rewrite $4$ as $2^2$ . Therefore, we have:

$4^x = (2^2)^x = 2^{2x}$

Since we know that $y = 2^x$ , we can express $2^{2x}$ in terms of $y$ :

$2^{2x} = (2^x)^2 = y^2$

Thus, we have:

$4^x = y^2$

Step 2

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Answer

Substituting $4^x$ with $y^2$ and $2^x$ with $y$ , we get:

$8(y^2) - 9(y) + 1 = 0$

This is a quadratic equation in the standard form $Ay^2 + By + C = 0$ , where:

We can apply the quadratic formula:

$y = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

Substituting the values:

$y = \frac{-(-9) \pm \sqrt{(-9)^2 - 4 \cdot 8 \cdot 1}}{2 \cdot 8}$

Calculating the discriminant:

$y = \frac{9 \pm \sqrt{81 - 32}}{16}$

$y = \frac{9 \pm \sqrt{49}}{16}$

$y = \frac{9 \pm 7}{16}$

Calculating the two possible values for $y$ :

Thus, the solutions for $y$ are $y = 1$ and $y = \frac{1}{8}$ .