Question

$if \: log_{2}(x) \times log_{2}( \frac{x}{16} ) + 4 = 0 \: then \: x \: is \: equal \: to$
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Answer 1

Answer:

$log_{2}(x) + log_{2}( \frac{x}{16} ) + 4 = 0 \\ = > log_{2}(x + \frac{x}{16} ) = - 4 \\ = > {2}^{ - 4} = x + \frac{x}{16} \\ = > \frac{1}{ {2}^{4} } = \frac{16x + x}{16} \\ = > \frac{1}{16} = \frac{17x}{16} \\ = > 17x = 1 \\ = > x = \frac{1}{17}$

the value of x is

1/17

.

Answer 2

Answer:

$\huge\bold {\underline{\underline{Solution}}}$

$\implies log_{2}(x) + log_{2} \left(\dfrac{x}{16} \right) + 4 = 0$

$\implies log_{2}\left(x +\dfrac{x}{16}\right) = - 4$

$\implies {2}^{ - 4} = x + \dfrac{x}{16}$

$\implies \dfrac{1}{ {2}^{4} } = \dfrac{16x + x}{16}$

$\implies \dfrac{1}{16} = \dfrac{17x}{16}$

$\implies17x = 1$

$\implies \: x = \dfrac{1}{17}$