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Theresa bought a house on 2 January 1970 for £8000 - AQA - A-Level Maths Pure - Question 8 - 2019 - Paper 2

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Theresa bought a house on 2 January 1970 for £8000. The house was valued by a local estate agent on the same date every 10 years up to 2010. The valuations are sho... show full transcript

Worked Solution & Example Answer:Theresa bought a house on 2 January 1970 for £8000 - AQA - A-Level Maths Pure - Question 8 - 2019 - Paper 2

Step 1

Show that $V = pq$ can be written as $\log_{10} V = \log_{10} p + \log_{10} q$.

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Answer

To express the valuation price VV in a logarithmic form, we begin with the equation:

V=pqV = pq

Taking the logarithm of both sides using base 10, we have:

log⁡10V=log⁡10(pq)\log_{10} V = \log_{10} (pq)

By applying the logarithmic property that states log⁡10(ab)=log⁡10a+log⁡10b\log_{10} (ab) = \log_{10} a + \log_{10} b, we can expand the equation:

log⁡10V=log⁡10p+log⁡10q\log_{10} V = \log_{10} p + \log_{10} q

This shows that the valuation price can indeed be expressed in terms of the logarithm of constants pp and qq.

Step 2

The values in the table of $\log_{10} V$ against $t$ have been plotted and a line of best fit has been drawn.

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Answer

From the graph, we can observe the relationship between tt and log⁡10V\log_{10} V. Using two points from the line of best fit, we can derive the linear equation:

  1. Calculate the slope:

    • Let points be (10,4.28)(10, 4.28) and (20,4.61)(20, 4.61).
    • Slope m=4.61−4.2820−10=0.3310=0.033m = \frac{4.61 - 4.28}{20 - 10} = \frac{0.33}{10} = 0.033.

  2. Using the point-slope form to obtain the linear equation:

    • Using point (10,4.28)(10, 4.28):
    • log⁡10V−4.28=0.033(t−10)\log_{10} V - 4.28 = 0.033(t - 10)

  3. Rearranging gives the equation of the line.

    • Finally, this presents the form of log⁡10V\log_{10} V as a linear function of tt.

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