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3. (a) Given that $$2 \log(4 - x) = \log(x + 8)$$ show that $$x^2 - 9x + 8 = 0$$ (b) (i) Write down the roots of the equation $$x^2 - 9x + 8 = 0$$ (ii) State which of the roots in (b)(i) is not a solution of $$2 \log(4 - x) = \log(x + 8)$$ giving a reason for your answer. - Edexcel - A-Level Maths Pure - Question 5 - 2020 - Paper 2

Question 5

$3.-(a)-Given-that--$$2-\log(4---x)-=-\log(x-+-8)$$--show-that--$$x^2---9x-+-8-=-0$$--(b)-(i)-Write-down-the-roots-of-the-equation--$$x^2---9x-+-8-=-0$$--(ii)-State-which-of-the-roots-in-(b)(i)-is-not-a-solution-of--$$2-\log(4---x)-=-\log(x-+-8)$$-giving-a-reason-for-your-answer.-Edexcel-A-Level Maths Pure-Question 5-2020-Paper 2.png$

3. (a) Given that $$2 \log(4 - x) = \log(x + 8)$$ show that $$x^2 - 9x + 8 = 0$$ (b) (i) Write down the roots of the equation $$x^2 - 9x + 8 = 0$$ (ii) State w... show full transcript

Worked Solution & Example Answer:3. (a) Given that $$2 \log(4 - x) = \log(x + 8)$$ show that $$x^2 - 9x + 8 = 0$$ (b) (i) Write down the roots of the equation $$x^2 - 9x + 8 = 0$$ (ii) State which of the roots in (b)(i) is not a solution of $$2 \log(4 - x) = \log(x + 8)$$ giving a reason for your answer. - Edexcel - A-Level Maths Pure - Question 5 - 2020 - Paper 2

Step 1

Given that $$2 \log(4 - x) = \log(x + 8)$$ show that $$x^2 - 9x + 8 = 0$$

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Answer

To show that the equation holds, we will start with the given logarithmic equation:

$2 \log(4 - x) = \log(x + 8)$

Using the properties of logarithms, we can rewrite this as:

$\log((4 - x)^2) = \log(x + 8)$

This implies:

$(4 - x)^2 = x + 8$

Expanding the left side gives:

$16 - 8x + x^2 = x + 8$

Rearranging all terms to one side results in:

$x^2 - 9x + 8 = 0$

Thus, we have shown that the statement is correct.

Step 2

(i) Write down the roots of the equation $$x^2 - 9x + 8 = 0$$

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Answer

To find the roots of the quadratic equation $x^2 - 9x + 8 = 0$ , we can factor the equation:

$(x - 1)(x - 8) = 0$

Setting each factor equal to zero provides the roots:

$x = 1$
$x = 8$

Step 3

(ii) State which of the roots in (b)(i) is not a solution of $$2 \log(4 - x) = \log(x + 8)$$ giving a reason for your answer.

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Answer

We have found the roots $x = 1$ and $x = 8$ . To determine which root is not a solution of the original logarithmic equation, we must evaluate:

For $x = 1$ : $\log(4 - 1) = \log(3)$ $\log(1 + 8) = \log(9)$ Thus, $2 \log(3) = \log(9)$ holds true.
For $x = 8$ : $\log(4 - 8) = \log(-4)$ Since the logarithm of a negative number is undefined, this means that $x = 8$ is not a valid solution for the original equation.

Therefore, the root $x = 8$ is not a solution.

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