SATHEE: Theory of Equations 1 Question 54

Theory of Equations 1 Question 54

55. Solve $2 \log_{x} a + \log_{a x} a + 3 \log_{b} a = 0$ , where $a > 0, b = a^{2} x$

(1978, 3M )

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Solution:

The given equation can be rewritten as

$\begin{array}{cc} \frac{2}{\log_{a} x} + \frac{1}{\log_{a} a x} + \frac{3}{\log_{a} a^{2} x} = 0 [∵ b = a^{2} x, given] \\ \Rightarrow & \frac{2}{\log_{a} x} + \frac{1}{1 + \log_{a} x} + \frac{3}{2 + \log_{a} x} = 0 \\ \Rightarrow & \frac{2}{t} + \frac{1}{1 + t} + \frac{3}{2 + t} = 0, where t = \log_{a} x \\ \Rightarrow & 2 (1 + t) (2 + t) + 3 t (1 + t) + t (2 + t) = 0 \\ \Rightarrow & 6 t^{2} + 11 t + 4 = 0 \\ \Rightarrow & (2 t + 1) (3 t + 4) = 0 \\ \Rightarrow & t = - \frac{1}{2} or - \frac{4}{3} \\ ∴ & \log_{a} x = - \frac{1}{2} or \log_{a} x = - \frac{4}{3} \\ \Rightarrow & x = a^{- 1 / 2} \\ or & x = a \end{array}$

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Theory of Equations 1 Question 54

55. Solve 2logx⁡a+logax⁡a+3logb⁡a=0, where a>0,b=a2x

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55. Solve $2 \log_{x} a + \log_{a x} a + 3 \log_{b} a = 0$ , where $a > 0, b = a^{2} x$