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2 log(x + a) = log(16a^6), where a is a positive constant - Edexcel - A-Level Maths Pure - Question 9 - 2017 - Paper 3
Question 9
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) log(9y + b) - log_2(2y - b) = 2, whe... show full transcript
Worked Solution & Example Answer:2 log(x + a) = log(16a^6), where a is a positive constant - Edexcel - A-Level Maths Pure - Question 9 - 2017 - Paper 3
Step 1
Find x in terms of a
Answer
To solve the equation, we start with the logarithmic expression:
- Apply the power rule of logarithms:
ightarrow ext{log}((x + a)^2) = ext{log}(16a^6)
2. Remove the logarithms by equating the arguments: $$(x + a)^2 = 16a^6$$ 3. Take the square root of both sides: $$x + a = 4a^3$$ 4. Solve for x: $$x = 4a^3 - a$$ 5. The final answer in simplest form is: $$x = a(4a^2 - 1)$$Step 2
Find y in terms of b
Answer
To solve for y in the given logarithmic equation:
-
Apply the properties of logarithms:
This simplifies to:
-
Exponentiate to remove the logarithm:
-
Cross-multiply to eliminate the fraction:
-
Expand and rearrange to isolate y:
ightarrow 9y + 101b = 200y
200y - 9y = 101b ightarrow 191y = 101b ightarrow y = \frac{101b}{191}
6. The final answer in simplest form is: $$y = \frac{101b}{191}$$