A-Level Edexcel Maths Pure Exponential & Logarithms: 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 2

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 2

Step 1

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

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Answer

To prove that ( a = \frac{b^2}{b - 1} ), we start with the given equation:

\text{log } a - \text{log } b = \text{log}(a - b)

Using the logarithmic identity, we can rewrite the left-hand side:

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

Taking exponentials on both sides results in:

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

Multiplying through by $b$ gives:

a = b(a - b)

Expanding the right-hand side, we have:

a = ab - b^2

Rearranging yields:

ab - a = b^2

Factoring out $a$ leads to:

a(b - 1) = b^2

Finally, dividing both sides by $(b - 1)$ yields:

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 restrictions on the value of $b$ are:

$b > 1$ : Since $a$ must be positive, and from our derived equation $a = \frac{b^2}{b - 1}$ , as $b$ approaches 1 from the right, $a$ approaches infinity, making $b$ less than or equal to 1 invalid as $a$ cannot be greater than or equal to $b$ for valid positive values.
Together with the condition $a > b$ and $a, b > 0$ , this leads us to restrict $b$ such that:

$b > 1$

This ensures that both $a$ and $b$ remain positive with $a$ always being greater than $b$ .

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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