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Given that $\log_{a} y = 2\log_{a} 7 + \log_{a} 4 + \frac{1}{2}$, find $y$ in terms of $a$ - AQA - A-Level Maths Mechanics - Question 7 - 2018 - Paper 3

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Given-that-$\log_{a}-y-=-2\log_{a}-7-+-\log_{a}-4-+-\frac{1}{2}$,-find-$y$-in-terms-of-$a$-AQA-A-Level Maths Mechanics-Question 7-2018-Paper 3.png

Given that $\log_{a} y = 2\log_{a} 7 + \log_{a} 4 + \frac{1}{2}$, find $y$ in terms of $a$. --- When asked to solve the equation $2\log_{9} x = \log_{9} 4 - \log_... show full transcript

Worked Solution & Example Answer:Given that $\log_{a} y = 2\log_{a} 7 + \log_{a} 4 + \frac{1}{2}$, find $y$ in terms of $a$ - AQA - A-Level Maths Mechanics - Question 7 - 2018 - Paper 3

Step 1

Find $y$ in terms of $a$

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Answer

Starting from the equation:

logay=2loga7+loga4+12\log_{a} y = 2\log_{a} 7 + \log_{a} 4 + \frac{1}{2}

Using the properties of logarithms, we can rewrite this as:

logay=loga(72)+loga4+loga(a1/2)\log_{a} y = \log_{a} (7^2) + \log_{a} 4 + \log_{a} (a^{1/2})

Combining the logarithms, we get:

logay=loga(494a1/2)\log_{a} y = \log_{a} (49 \cdot 4 \cdot a^{1/2})

This implies:

y=494a1/2=196ay = 49 \cdot 4 \cdot a^{1/2} = 196 \sqrt{a}

Step 2

Explain what is wrong with the student's solution.

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Answer

The error in the student's solution stems from accepting 32\frac{3}{2} as a valid solution without considering the domain of the logarithm. Logarithmic functions require positive inputs, so xx must be greater than zero. Since xx cannot be 32\frac{-3}{2}, this solution should be rejected. The solution 32\frac{3}{2} is acceptable, but the student's context misrepresents that as a valid output. Therefore, the correct interpretation is that 32\frac{3}{2} is a potential solution, but the negative value must be rejected due to the logarithmic constraints.

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