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Basics & Logarithms: Denotes Product Values Satisfying Equation Value

JEE Maths question with a full step-by-step solution.

Question
If pp denotes the product of the two values of xx satisfying the equation
log⁡2x(2)+log⁡4(2x)=−32\log_{2x}(2) + \log_4(2x) = -\frac{3}{2}
 then the value of log⁡2 ⁣(1p)\log_2\!\left(\dfrac{1}{p}\right) is
Solution
Answer: 5
Step 1: Let y=log⁡2(2x)y = \log_2(2x). Then:
log⁡2x(2)=1log⁡2(2x)=1y,log⁡4(2x)=log⁡2(2x)2=y2\log_{2x}(2) = \frac{1}{\log_2(2x)} = \frac{1}{y}, \qquad \log_4(2x) = \frac{\log_2(2x)}{2} = \frac{y}{2}
 Step 2: The equation becomes:
1y+y2=−32  ⟹  2+y22y=−32  ⟹  y2+3y+2=0  ⟹  (y+2)(y+1)=0\frac{1}{y} + \frac{y}{2} = -\frac{3}{2} \implies \frac{2+y^2}{2y} = -\frac{3}{2} \implies y^2+3y+2=0 \implies (y+2)(y+1)=0
 y=−2y = -2 or y=−1y = -1. Step 3: Recover xx. y=−2y=-2: log⁡2(2x)=−2⇒2x=14⇒x=18\log_2(2x)=-2 \Rightarrow 2x=\dfrac{1}{4} \Rightarrow x=\dfrac{1}{8} y=−1y=-1: log⁡2(2x)=−1⇒2x=12⇒x=14\log_2(2x)=-1 \Rightarrow 2x=\dfrac{1}{2} \Rightarrow x=\dfrac{1}{4} Both have 2x>02x > 0 and base ≠1\neq 1. Both valid. Step 4: p=18⋅14=132p = \dfrac{1}{8} \cdot \dfrac{1}{4} = \dfrac{1}{32}.
log⁡2 ⁣(1p)=log⁡2(32)=5\log_2\!\left(\frac{1}{p}\right) = \log_2(32) = 5
 Answer: 5
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