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Question Number 74582 by Maclaurin Stickker last updated on 26/Nov/19

Find all values of x: (2^x )^(x^2 −8) =32

$${Find}\:{all}\:{values}\:{of}\:{x}: \\ $$$$\left(\mathrm{2}^{{x}} \right)^{{x}^{\mathrm{2}} −\mathrm{8}} =\mathrm{32} \\ $$

Answered by MJS last updated on 26/Nov/19

(a^b )^c =a^(bc) 2^(x^3 −8x) =2^5 x^3 −8x−5=0 this has 3 real solutions ⇒ we need the trigonometric formula x_1 =((4(√6))/3)sin ((π/3)+(1/3)arcsin ((15(√6))/(64))) ≈3.10043 x_2 =−((4(√6))/3)sin ((π/6)+(1/3)arcsin ((15(√6))/(64))) ≈−2.43931 x_3 =−((4(√6))/3)sin ((1/3)arcsin ((15(√6))/(64))) ≈−.661120

$$\left({a}^{{b}} \right)^{{c}} ={a}^{{bc}} \\ $$$$\mathrm{2}^{{x}^{\mathrm{3}} −\mathrm{8}{x}} =\mathrm{2}^{\mathrm{5}} \\ $$$${x}^{\mathrm{3}} −\mathrm{8}{x}−\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{has}\:\mathrm{3}\:\mathrm{real}\:\mathrm{solutions}\:\Rightarrow\:\mathrm{we}\:\mathrm{need}\:\mathrm{the} \\ $$$$\mathrm{trigonometric}\:\mathrm{formula} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{3}}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{15}\sqrt{\mathrm{6}}}{\mathrm{64}}\right)\:\approx\mathrm{3}.\mathrm{10043} \\ $$$${x}_{\mathrm{2}} =−\frac{\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{3}}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{15}\sqrt{\mathrm{6}}}{\mathrm{64}}\right)\:\approx−\mathrm{2}.\mathrm{43931} \\ $$$${x}_{\mathrm{3}} =−\frac{\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{3}}\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{15}\sqrt{\mathrm{6}}}{\mathrm{64}}\right)\:\approx−.\mathrm{661120} \\ $$

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