Question

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Find the domain of the function

$f(x) = \sqrt[6]{ {4}^{x} + {8}^{ \frac{2x}{3} } - {2}^{2(x - 1)} - 52}$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = \sqrt[6]{ {4}^{x} + {\bigg[8\bigg]}^{ \dfrac{2x}{3} } - {2}^{2(x - 1)} - 52}$

Domain is defined as set of those values of x for which function is well defined.

So, f(x) is defined when

$\rm :\longmapsto\: {4}^{x} + {\bigg[8\bigg]}^{ \dfrac{2x}{3} } - {2}^{2(x - 1)} - 52 \geqslant 0$

can be rewritten as

$\rm :\longmapsto\: {2}^{2x} + {\bigg[ {2}^{3} \bigg]}^{ \dfrac{2x}{3} } - {2}^{2(x - 1)} \geqslant 52$

$\rm :\longmapsto\: {2}^{2x} + {2}^{2x} - {2}^{2(x - 1)} \geqslant 52$

$\rm :\longmapsto\: {2}^{2x} + {2}^{2x} - {2}^{2x - 2} \geqslant 52$

$\rm :\longmapsto\: {2}^{2x} + {2}^{2x} - {2}^{2x} \times {2}^{ - 2} \geqslant 52$

$\rm :\longmapsto\: {2}^{2x} + {2}^{2x} - \dfrac{ {2}^{2x} }{4} \geqslant 52$

$\rm :\longmapsto\: \dfrac{4. {2}^{2x} + 4. {2}^{2x} - {2}^{2x} }{4} \geqslant 52$

$\rm :\longmapsto\: \dfrac{7. {2}^{x} }{4} \geqslant 52$

$\rm :\implies\: {2}^{2x} \geqslant \dfrac{208}{7}$

$\rm :\implies\: {4}^{x} \geqslant \dfrac{208}{7}$

We know, that

$\boxed{ \tt{ \: {x}^{ log_{x}(y) } = y \: \: }}$

So, using this, we get

$\rm :\implies\: {4}^{x} \geqslant {4}^{ log_{4}\bigg[\dfrac{208}{7} \bigg]}$

$\bf\implies \:x \geqslant log_{4}\bigg[\dfrac{208}{7} \bigg]$

Answer 2

Answer:

$⟹x⩾log4[7208]$