Find the value of x if $2 {log}_{x} 7 + {log}_{7 x} 7 + 3 {log}_{49 x} 7 = 0$

x = \frac{- 4}{3}

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x = 7^{- 1 / 2}

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x = 7^{- 4 / 3}

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Either (B) or (C)

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Solution

The correct option is D Either (B) or (C)
Given
$2 l o g_{x} 7 + l o g_{7 x} 7 + 3 l o g_{49 x} 7 = 0$ $

$2 \frac{l o g_{7} 7}{l o g_{7} x} + \frac{l o g_{7} 7}{l o g_{7} 7 x} + 3 \frac{l o g_{7} 7}{l o g_{7} 7} = 0$

$2 \frac{1}{l o g_{7} x} + \frac{1}{l o g_{7} 7 x} + 3 \frac{1}{l o g_{7} (49 x)} = 0$

$\frac{2}{l o g_{7} x} + \frac{1}{l o g_{7} 7 + l o g_{7} x} + 3 \frac{1}{l o g_{7} 49 x} = 0$

$\frac{2}{l o g_{7} x} + \frac{1}{1 + l o g_{7} x} + \frac{3}{2 + l o g_{7} x} = 0$

Let $l o g_{7} x = a$

$\frac{2}{a} + \frac{1}{1 + a} + \frac{3}{2 + a} = 0$

By taking LCM and cross multiplication

$2 (1 + a) (2 + a) + a (2 + a) + 3 a (1 + a) = 0$

$2 (a^{2} + 3 a + 2) + (2 a + a^{2}) + (3 a + 3 a^{2}) = 0$

$6 a^{2} + 11 a + 4 = 0$

$(3 a + 4) (2 a + 1) = 0$

$3 a = - 4$

$a = - \frac{4}{3}$

$l o g_{7} x = - \frac{4}{3}$

$x = 7^{- \frac{4}{3}}$

If $2 a + 1 = 0$

$2 a = - 1$

$a = - \frac{1}{2}$

$l o g_{7} x = - \frac{1}{2}$

$x = 7^{- \frac{1}{2}}$
Hence the values of $x$ are $7^{- \frac{1}{2}}$ and $7^{\frac{- 4}{3}}$

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