Basics & LogarithmshardFree

Basics & Logarithms: Solutions Equation Value

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

Question

x_1

and

x_2

are the two solutions of the equation

3^{\log_2 x} - 12 \cdot x^{\log_{16} 9} + 27 = 0

, then the value of

x_1^2 + x_2^2

Solution

Answer: 272

Step 1: Simplify

\log_{16} 9 = \dfrac{\log_2 9}{\log_2 16} = \dfrac{2\log_2 3}{4} = \dfrac{\log_2 3}{2}

. So

x^{\log_{16} 9} = x^{(\log_2 3)/2} = \sqrt{x^{\log_2 3}} = \sqrt{3^{\log_2 x}}

(using

a^{\log_b c} = c^{\log_b a}

). Step 2: Let

u = 3^{\log_2 x} > 0

. The equation becomes:

u - 12\sqrt{u} + 27 = 0

Let

v = \sqrt{u}

v^2 - 12v + 27 = 0 \implies (v-3)(v-9) = 0 \implies v=3

v=9

. Step 3: Recover

x

v=3 \Rightarrow u=9 \Rightarrow 3^{\log_2 x} = 9 = 3^2 \Rightarrow \log_2 x = 2 \Rightarrow x_1 = 4

v=9 \Rightarrow u=81 \Rightarrow 3^{\log_2 x} = 81 = 3^4 \Rightarrow \log_2 x = 4 \Rightarrow x_2 = 16

Step 4:

x_1^2 + x_2^2 = 16 + 256 = 272

. Answer: 272

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