Find the value of x which satisfies the relation:

$l o g_{10} 3 + l o g_{10} 4 x + 1 = l o g_{10} x + 1 + 1$

$\frac{2}{3}$

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$\frac{7}{2}$

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$\frac{3}{2}$

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$1$

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Solution

The correct option is B
$\frac{7}{2}$

$l o g_{10} 3 + l o g_{10} 4 x + 1 = l o g_{10} x + 1 + l o g_{10}^{10}$

$3 (4 x + 1) = 10 (x + 1)$

$12 x + 3 = 10 x + 10$

$2 x = 7$

$x = \frac{7}{2}$

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