A-Level Edexcel Maths Pure Inequalities: Given that $a > b

Given that $a > b > 0$ and that $a$ and $b$ satisfy the equation $$ olimits ext{log} \, a - ext{log} \, b = ext{log} \, (a - b)$$ (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 10 - 2019 - Paper 1

Question 10

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

Worked Solution & Example Answer:Given that $a > b > 0$ and that $a$ and $b$ satisfy the equation $$ olimits ext{log} \, a - ext{log} \, b = ext{log} \, (a - b)$$ (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 10 - 2019 - Paper 1

Step 1

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

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Answer

Starting from the given equation:

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

Using the properties of logarithms:

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

We can equate the arguments of the logarithms, leading us to:

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

Now, multiply both sides by $b$ :

$a = b(a - b)$

Expanding the right-hand side gives:

$a = ab - b^2$

Rearranging this results in:

$ab - a = b^2$

Factoring out $a$ from the left side yields:

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

Finally, solve for $a$ :

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

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:

$b > 1$

This restriction is necessary because:

From the derived equation for $a$ , we have $b - 1$ as the denominator. Therefore, to avoid division by zero, $b$ cannot equal 1.
Additionally, since $b$ must be less than $a$ (as given), we need $b$ to be strictly greater than 1 for $a$ to remain positive; thus ensuring it adheres to the condition $a > b > 0$ .

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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