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The curve with equation $y = 3x^2$ meets the curve with equation $y = 15 - 2^{-x}$ at the point P - Edexcel - A-Level Maths Pure - Question 7 - 2020 - Paper 2

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The curve with equation $y = 3x^2$ meets the curve with equation $y = 15 - 2^{-x}$ at the point P. Find, using algebra, the exact x coordinate of P.

Worked Solution & Example Answer:The curve with equation $y = 3x^2$ meets the curve with equation $y = 15 - 2^{-x}$ at the point P - Edexcel - A-Level Maths Pure - Question 7 - 2020 - Paper 2

Step 1

Combine the Equations

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Answer

To find the x-coordinate of point P, we begin by setting the equations equal to each other: 3x2=152x3x^2 = 15 - 2^{-x} This can be rearranged to: 2x=153x22^{-x} = 15 - 3x^2

Step 2

Rearranging to Form a Power Equation

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Answer

Next, we can rewrite the equation in terms of powers of 2: 153x2=2x15 - 3x^2 = 2^{-x} To eliminate the negative exponent, we can rearrange the equation to: 15 - 3x^2 = rac{1}{2^x}

Step 3

Multiplying Through by 2^x

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Answer

Multiply both sides by 2x2^x: 2x(153x2)=12^x (15 - 3x^2) = 1 Which leads to: 15imes2x3x2imes2x=115 imes 2^x - 3x^2 imes 2^x = 1

Step 4

Logarithmic Transformation

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Answer

From the equation 3x2=152x3x^2 = 15 - 2^{-x}, we can isolate 2x2^{-x} and switch to a logarithmic form: 2^x = rac{15}{3} Thus, we find: x=extlog2(3)x = - ext{log}_2(3) The exact x-coordinate of P is: x=extlog32x = ext{log}_3 2

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