A-Level Edexcel Maths Pure Modelling with 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) ... 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 prove the equation, we start from:

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

Using the logarithmic identity, we can rewrite this as:

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

This leads us to:

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

Multiplying both sides by $b$ gives us:

$a = b(a - b)$

Rearranging yields:

$a - ab = -b^2$

Thus,

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

Now, solving for $a$ results in:

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

Since $b$ must be less than 1 to avoid division by zero, we adjust this to:

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

Step 2

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

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Answer

The full restriction on the value of $b$ is that $b$ must be less than 1, specifically:

$b < 1$

This restriction arises because when substituting $b$ into the derived equation for $a$ , if $b$ equals or exceeds 1, it would result in a denominator of zero or a negative, which is not permissible for the logarithm (as $a$ must remain positive). Thus, we also have that $b > 0$ for the original problem's condition, leading to the restriction:

$0 < b < 1$ .

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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