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Question Number 172085 by Mikenice last updated on 23/Jun/22
solve  (3^(x^3 −72x+39) −9(√3))×log(7−x)=0
$${solve} \\ $$$$\left(\mathrm{3}^{{x}^{\mathrm{3}} −\mathrm{72}{x}+\mathrm{39}} −\mathrm{9}\sqrt{\mathrm{3}}\right)×{log}\left(\mathrm{7}−{x}\right)=\mathrm{0} \\ $$
Answered by mr W last updated on 23/Jun/22
log (7−x)=0 ⇒7−x=1 ⇒x=6   or  3^(x^3 −72x+39) −9(√3)=0  3^(x^3 −72x+39) =9(√3)=3^(2+(1/2))   x^3 −72x+39=2+(1/2)  x^3 −72x+((73)/2)=0  ⇒x=4(√6) sin ((1/3) sin^(−1) ((73(√6))/(1152))+((2kπ)/3)) (k=0,1,2)
$$\mathrm{log}\:\left(\mathrm{7}−{x}\right)=\mathrm{0}\:\Rightarrow\mathrm{7}−{x}=\mathrm{1}\:\Rightarrow{x}=\mathrm{6}\: \\ $$$${or} \\ $$$$\mathrm{3}^{{x}^{\mathrm{3}} −\mathrm{72}{x}+\mathrm{39}} −\mathrm{9}\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$$\mathrm{3}^{{x}^{\mathrm{3}} −\mathrm{72}{x}+\mathrm{39}} =\mathrm{9}\sqrt{\mathrm{3}}=\mathrm{3}^{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${x}^{\mathrm{3}} −\mathrm{72}{x}+\mathrm{39}=\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} −\mathrm{72}{x}+\frac{\mathrm{73}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{4}\sqrt{\mathrm{6}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{73}\sqrt{\mathrm{6}}}{\mathrm{1152}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)\:\left({k}=\mathrm{0},\mathrm{1},\mathrm{2}\right) \\ $$

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