Question

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Product of all values of x satisfying the equation


$\sqrt{2 {}^{x} \sqrt[3]{ {4}^{x} (0.125) {}^{1 \div x} } } = 4( \sqrt[3]{2} )$

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Answer 1

Answer :

Given,

$\sqrt{ {2}^{x} \sqrt[3]{ {4}^{x} {(0.125)}^{1 \div x} } } = 4( \sqrt[3]{2}) \\ \\ \implies {( {2}^{x} \sqrt[3]{ {4}^{x} {(0.125)}^{1 \div x} } )}^{ \frac{1}{2} } = {2}^{2} \times {2}^{ \frac{1}{3} } = {2}^{ \frac{7}{3} }$

Now, squaring both sides, we get

${2}^{x} \sqrt[3]{ {4}^{x} {(0.125)}^{1 \div x} } = {2}^{ \frac{14}{3} } \\ \\ \implies \sqrt[3]{ {4}^{x} {(0.125)}^{1 \div x} } = {2}^{ \frac{14}{3} } \times {2}^{ - x} = \ {2}^{ \frac{14 - 3x}{3} } \\ \\ \implies {( {4}^{x} {(0.125)}^{1 \div x}) }^{ \frac{1}{3} } = {2}^{ \frac{14 - 3x}{3} }$

Cubing both sides, we get

${4}^{x} {(0.125)}^{1 \div x} = {2}^{14 - 3x} \\ \\ \implies {0.125}^{1 \div x} = {2}^{14 - 3x} \times {4}^{ - x} \\ \\ \implies {0.125}^{1 \div x} = {2}^{14 - 3x} \times {2}^{ - 2x} \\ \\ \implies {0.125}^{ \frac{1}{x} } = {2}^{14 - 5x}$

Now, taking x as a power, we get

$0.125 = {2}^{x(14 - 5x)}$

Taking (log) to both sides, we get

$log(0.125) = (14x - 5 {x}^{2} )(log2) \\ \\ \implies log {(0.5)}^{3} = (14x - 5 {x}^{2} )(log2) \\ \\ \implies 3 \: log(0.5) = (14x - 5 {x}^{2} )(log2) \\ \\ \implies 3 \: log( \frac{1}{2} ) = (14x - 5 {x}^{2} )(log2) \\ \\ \implies - 3(log2) = (14x - 5 {x}^{2} )(log2)$

{ Since, log$(\frac{1}{2})$ = log1 - log2 = - log2 }

$\implies - 3 = 14x - 5 {x}^{2} \\ \\ \implies 5 {x}^{2} - 14x - 3 = 0 \\ \\ \implies 5 {x}^{2} - 15x + x - 3 = 0 \\ \\ \implies 5x(x - 3) + 1(x - 3) = 0 \\ \\ \implies (x - 3)(5x + 1) = 0$

So, the values of x are 3, $-\frac{1}{5}$ and they satisfy the given equation.

Thus, the product of the values of x is

= $3\times (-\frac{1}{5})$

= $-\frac{3}{5}$

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Answer 2

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