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Given that $ ext{log}_g y = 2 ext{log}_g 7 + ext{log}_g 4 + rac{1}{2}$, find $y$ in terms of $a$ - AQA - A-Level Maths Pure - Question 7 - 2018 - Paper 3

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Question 7

Given-that-$-ext{log}_g-y-=-2-ext{log}_g-7-+--ext{log}_g-4-+--rac{1}{2}$,-find-$y$-in-terms-of-$a$-AQA-A-Level Maths Pure-Question 7-2018-Paper 3.png

Given that $ ext{log}_g y = 2 ext{log}_g 7 + ext{log}_g 4 + rac{1}{2}$, find $y$ in terms of $a$. When asked to solve the equation $2 ext{log}_g x = ext{log}_9 ... show full transcript

Worked Solution & Example Answer:Given that $ ext{log}_g y = 2 ext{log}_g 7 + ext{log}_g 4 + rac{1}{2}$, find $y$ in terms of $a$ - AQA - A-Level Maths Pure - Question 7 - 2018 - Paper 3

Step 1

Find $y$ in terms of $a$

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Answer

We start with the equation:

ext{log}_g y = 2 ext{log}_g 7 + ext{log}_g 4 + rac{1}{2}

Using properties of logarithms, we can combine the log terms:

ext{log}_g y = ext{log}_g(7^2) + ext{log}_g(4) + rac{1}{2}

Now, applying the product property of logarithms:

ext{log}_g y = ext{log}_g(49) + ext{log}_g(4) + rac{1}{2} ext{log}_g y = ext{log}_g(196) + rac{1}{2}

We can convert rac{1}{2} into logarithmic form:

rac{1}{2} = ext{log}_g( ext{sqrt}(g))

Thus, we can represent this as:

extloggy=extlogg(196imesextsqrt(g)) ext{log}_g y = ext{log}_g(196 imes ext{sqrt}(g))

Hence,

y=196extsqrt(g)y = 196 ext{sqrt}(g)

Step 2

Explain what is wrong with the student's solution

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Answer

The student's solution incorrectly concludes that x = rac{3}{2} is valid. However, the value rac{3}{2} should be rejected because it is not possible to evaluate ext{log}_g rac{3}{2}. Since this value is derived from determining the logarithm of a fraction, this specific case leads to the conclusion that the logarithm is not defined for this input.

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