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Find the value of x
$\log (2 x) - \log (2^{x}) = \log (x^{2}) - \log (x)$

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2

1

3

- 1

answer is B.

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Detailed Solution

\log (2 x) - \log (2^{x}) = \log (x^{2}) - \log (x)

\log (\frac{2 x}{2^{x}}) = \log (\frac{x^{2}}{x})

\frac{2 x}{2^{x}} = \frac{x^{2}}{x}

(taking antilog on both sides)

2 x^{2} = x^{2} 2^{x}

(cross multiply)

2 x^{2} - x^{2} 2^{x} = 0

x^{2} ({2 - 2}^{x}) = 0

x^{2} = 0

x = 0

(not valid)

({2 - 2}^{x}) = 0

{2 = 2}^{x}

{2^{1} = 2}^{x}

x = 1

∴ x = 1

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