A-Level Edexcel Maths Pure Arithmetic Sequences & Series: Given that $a > b

Given that $a > b > 0$ and that $a$ and $b$ satisfy the equation $$ ext{log } a - ext{log } b = ext{log}(a - b)$$ (a) show that $$a = \frac{b^2}{b - 1}$$ (b) Write down the full restriction on the value of $b$, explaining the reason for this restriction. - Edexcel - A-Level Maths Pure - Question 11 - 2019 - Paper 1

Question 11

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Given that $a > b > 0$ and that $a$ and $b$ satisfy the equation $$ ext{log } a - ext{log } b = ext{log}(a - b)$$ (a) show that $$a = \frac{b^2}{b - 1}$$ (b) W... show full transcript

Worked Solution & Example Answer:Given that $a > b > 0$ and that $a$ and $b$ satisfy the equation $$ ext{log } a - ext{log } b = ext{log}(a - b)$$ (a) show that $$a = \frac{b^2}{b - 1}$$ (b) Write down the full restriction on the value of $b$, explaining the reason for this restriction. - Edexcel - A-Level Maths Pure - Question 11 - 2019 - Paper 1

Step 1

show that $a = \frac{b^2}{b - 1}$

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Answer

To solve for $a$ , start with the equation:

$ext{log } a - ext{log } b = ext{log}(a - b).$

Using the log properties, this can be rewritten as:

$\text{log} \left( \frac{a}{b} \right) = \text{log}(a - b).$

This implies:

$\frac{a}{b} = a - b.$

By multiplying both sides by $b$ , we have:

$a = b(a - b) = ab - b^2.$

Rearranging gives:

$a - ab + b^2 = 0.$

Factoring out $a$ , we get:

$a(1 - b) = b^2.$

Thus,

$a = \frac{b^2}{1 - b}.$

Since $1 - b$ can be rewritten as $-(b - 1)$ , we have:

$$a = \frac{b^2}{b - 1}.$

Step 2

Write down the full restriction on the value of $b$, explaining the reason for this restriction.

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Answer

The restriction on the value of $b$ is such that:

$b > 1$ : This ensures that $a$ is positive because we need $a = \frac{b^2}{b - 1}$ , which requires that $b - 1 > 0$ .
$b$ can’t be equal to 1: If $b = 1$ , the expression becomes undefined as we would be dividing by zero. Thus, both conditions must hold true for the validity of the equation and to keep $a$ positive.

show that $a = \frac{b^2}{b - 1}$

Write down the full restriction on the value of $b$, explaining the reason for this restriction.

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