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Question Number 33200 by pieroo last updated on 12/Apr/18

$$\mathrm{Solve}{2}^{3\mathrm{n}+2}-7\times 2^{2\mathrm{n}+2}-31\times 2^{\mathrm{n}}-8=0,\mathrm{n}\in \mathbf{R}.$$$$\mathrm{I}\mathrm{need}\mathrm{some}\mathrm{help}\mathrm{with}\mathrm{this}$$

Answered by MJS last updated on 12/Apr/18

$$2^{3n+2}=4\times 8^n=4\times {\left(2^n\right)}^3$$$$-7\times 2^{2n+2}=-28\times 4^n=-28\times {\left(2^n\right)}^2$$$$x=2^n$$$$4x^3-28x^2-31x-8=0$$$$x_1{x}_2{x}_3=-8\Rightarrow \mathrm{try}\pm 1;\pm 2;\pm 4;\pm 8$$$$\Rightarrow x_1=8$$$$\frac{4x^3-28x^2-31x-8}{x-8}=4x^2+4x+1$$$$4x^2+4x+1=0$$$$x^2+x+\frac{1}{4}=0$$$${(x+\frac{1}{2})}^2=0\Rightarrow x_2=x_3=-\frac{1}{2}$$$$x=8\Rightarrow 2^n=8\Rightarrow n=3$$$$x=-\frac{1}{2}\Rightarrow 2^n=-\frac{1}{2}\Rightarrow \mathrm{no}\mathrm{solution}\mathrm{in}\mathbb{R}$$

Commented by pieroo last updated on 13/Apr/18

$$\mathrm{thanks}\mathrm{very}\mathrm{much}\mathrm{boss}$$

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