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log_x a +- log_{ax}a + 3log_{a^{2}x} a = 0

Question

$l o g_{x} a + - l o g_{a x} a + 3 l o g_{a^{2} x} a = 0$

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l o g_{x} a + - l o g_{a x} a + 3 l o g_{a^{2} x} a = 0

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Text solutionVerified

lo g_{x} a + (- lo g_{a x} a) + 3 lo g_{a^{2} x} a = 0

lo g_{x} a - lo g_{a x} a + 3 lo g_{a^{2} x} a = 0

\frac{1}{lo g _{a} x} - \frac{1}{lo g _{a} a x} + \frac{3}{lo g _{a} a ^{2} x} = 0

\frac{1}{lo g _{a} x} - \frac{1}{lo g _{a} a + lo g _{a} x} + \frac{3}{lo g _{a} a ^{2} + lo g _{a} x} = 0

\frac{1}{lo g _{a} x} - \frac{1}{1 + lo g _{a} x} + \frac{3}{2 + lo g _{a} x} = 0

Let

y = lo g_{a} x

∴ \frac{1}{y} - \frac{1}{1 + y} + \frac{3}{2 + y} = 0

\Rightarrow \frac{( 1 + y ) ( 2 + y ) - y ( 2 + y ) + 3 y ( 1 + y )}{y ( 1 + y ) ( 2 + y )} = 0

\Rightarrow (1 + y) (2 + y) - y (2 + y) + 3 y (1 + y) = 0

\Rightarrow (2 + y + 2 y + y^{2}) - (2 y + y^{2}) + (3 y + 3 y^{2}) = 0

\Rightarrow 2 + 3 y + y^{2} - 2 y - y^{2} + 3 y + 3 y^{2} = 0

\Rightarrow 3 y^{2} + 4 y + 2 = 0

Now by quadratic formula, we have

y = \frac{- 4 \pm 16 - 24}{6}

\Rightarrow y = \frac{- 2 \pm 2 2 i}{3}

Therefore,

lo g_{a} x = \frac{- 2 \pm 2 2 i}{3}

\Rightarrow x = a^{\frac{- 2 \pm 2 2 i}{3}}

Hence

x = a^{\frac{- 2 \pm 2 2 i}{3}}

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