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2 log(x + a) = log(16a^6), where a is a positive constant Find x in terms of a, giving your answer in its simplest form - Edexcel - A-Level Maths Pure - Question 9 - 2017 - Paper 3

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2-log(x-+-a)-=-log(16a^6),-where-a-is-a-positive-constant--Find-x-in-terms-of-a,-giving-your-answer-in-its-simplest-form-Edexcel-A-Level Maths Pure-Question 9-2017-Paper 3.png

2 log(x + a) = log(16a^6), where a is a positive constant Find x in terms of a, giving your answer in its simplest form. (3) i) log(9y + b) - log(2y - b) = 2, wh... show full transcript

Worked Solution & Example Answer:2 log(x + a) = log(16a^6), where a is a positive constant Find x in terms of a, giving your answer in its simplest form - Edexcel - A-Level Maths Pure - Question 9 - 2017 - Paper 3

Step 1

Find x in terms of a

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Answer

To solve for xx in the equation 2log(x+a)=log(16a6)2 \log(x + a) = \log(16a^6):

  1. Apply the Power Rule: Rewrite the left side: log((x+a)2)=log(16a6)\log((x + a)^2) = \log(16a^6)

  2. Remove the Logarithms: Set the arguments equal: (x+a)2=16a6(x + a)^2 = 16a^6

  3. Take the Square Root: Taking the square root of both sides gives: x+a=4a3x + a = 4a^3

  4. Isolate x: Solving for xx, we get: x=4a3ax = 4a^3 - a This simplifies to: x=a(4a21)x = a(4a^2 - 1)

Step 2

Find y in terms of b

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Answer

To solve for yy in the equation log(9y+b)log(2yb)=2\log(9y + b) - \log(2y - b) = 2:

  1. Apply the Quotient Rule for Logarithms: Combine the logarithms: log(9y+b2yb)=2\log\left(\frac{9y + b}{2y - b}\right) = 2

  2. Remove the Logarithm: Exponentiate both sides: 9y+b2yb=100\frac{9y + b}{2y - b} = 100

  3. Cross-Multiply: This leads to: 9y+b=100(2yb)9y + b = 100(2y - b)

  4. Expand and Rearrange: Expanding gives: 9y+b=200y100b9y + b = 200y - 100b Now rearranging terms results in: 9y+101b=200y9y + 101b = 200y Then, 101b=200y9y101b = 200y - 9y

  5. Isolate y: Finally, y(2009)=101by(200 - 9) = 101b Leading to: y=101b191y = \frac{101b}{191}

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