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5. (a) Find the positive value of $x$ such that \[ \log_x 64 = 2 \] (b) Solve for $x$ \[ \log_2 (11 - 6x) = 2 \log_2 (x - 1) + 3 \] - Edexcel - A-Level Maths: Pure - Question 6 - 2010 - Paper 4
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5. (a) Find the positive value of $x$ such that \[ \log_x 64 = 2 \] (b) Solve for $x$ \[ \log_2 (11 - 6x) = 2 \log_2 (x - 1) + 3 \]
Worked Solution & Example Answer:5. (a) Find the positive value of $x$ such that \[ \log_x 64 = 2 \] (b) Solve for $x$ \[ \log_2 (11 - 6x) = 2 \log_2 (x - 1) + 3 \] - Edexcel - A-Level Maths: Pure - Question 6 - 2010 - Paper 4
Step 1
Step 2
Solve for $x$\log_2 (11 - 6x) = 2 \log_2 (x - 1) + 3
Answer
First, we simplify the right side:
[ \log_2(11 - 6x) = \log_2((x - 1)^2) + 3 ]
Using the property of logarithms that states ( \log_b a + c = \log_b(a \cdot b^c) ), we can rewrite this as:
[ \log_2(11 - 6x) = \log_2((x - 1)^2 \cdot 2^3) ]
This leads to:
[ 11 - 6x = (x - 1)^2 \cdot 8 ]
Expanding the right side, we have:
[ 11 - 6x = 8(x^2 - 2x + 1) ]
This simplifies to:
[ 11 - 6x = 8x^2 - 16x + 8 ]
Rearranging gives:
[ 0 = 8x^2 - 10x - 3 ]
Now we can solve this quadratic equation using the quadratic formula:
[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 8 \cdot (-3)}}{2 \cdot 8} ]
Calculating the discriminant:
[ (-10)^2 - 4 \cdot 8 \cdot (-3) = 100 + 96 = 196 ]
Using this in the formula:
[ x = \frac{10 \pm 14}{16} ]
Thus, we find two values:
[ x = \frac{24}{16} = 1.5 \quad \text{and} \quad x = \frac{-4}{16} = -0.25 ]
Since we are looking for positive , we take ( x = 1.5 ).