A-Level Edexcel Maths Pure Equation of a Straight Line: Given that $y = 2^x$, (a) expr

Given that $y = 2^x$, (a) express $4^x$ in terms of $y$ - Edexcel - A-Level Maths Pure - Question 8 - 2015 - Paper 1

Question 8

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 8 - 2015 - Paper 1

Step 1

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Answer

Since we are given that $y = 2^x$ , we can express $4^x$ as follows:

Recall that $4$ can be rewritten as $2^2$ . Thus:

4^x &= (2^2)^x \\ &= 2^{2x}. \end{align*}$$ Now, using our expression for $y$, we can substitute $y = 2^x$: $$2^{2x} = (2^x)^2 = y^2.$$ Therefore, $$4^x = y^2.$$

Step 2

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Answer

Substituting $4^x$ from part (a) into the equation:

$8(4^x) - 9(2^x) + 1 = 0$

becomes:

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

This is a quadratic equation in terms of $y$ . We can solve for $y$ using the quadratic formula, where $a = 8$ , $b = -9$ , and $c = 1$ :

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

Calculating the discriminant:

$b^2 - 4ac = 81 - 32 = 49,$

we find:

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

This gives us two possible values for $y$ :

Since $y = 2^x$ , we solve for $x$ :

Thus the solutions for $x$ are:

$x = 0 \text{ or } x = -3.$