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Question Number 175121 by Michaelfaraday last updated on 20/Aug/22

Answered by BaliramKumar last updated on 20/Aug/22

$${\left(3^{-1}\right)}^{\frac{x^2-2x}{16-2x^3}}={\left(3^2\right)}^{\frac{1}{4x}}$$$$3^{\left(\frac{x^2-2x}{2x^3-16}\right)}=3^{\left(\frac{2}{4x}\right)}$$$$\frac{x^2-2x}{2(x^3-8)}=\frac{2}{4x}$$$$\frac{x^2-2x}{x^3-8}=\frac{1}{x}$$$$x^3-8=x^3-2x^2$$$$2x^2=8$$$$x=\pm 2$$

Commented by Frix last updated on 20/Aug/22

$$+2\mathrm{is}\mathrm{wrong}\mathrm{because}\frac{x^2-2x}{16-2x^3}\mathrm{is}\mathrm{not}\mathrm{defined}$$

Commented by Tawa11 last updated on 20/Aug/22

$$\mathrm{Great}\mathrm{sirs}$$

Commented by peter frank last updated on 24/Aug/22

$$\mathrm{good}$$

Answered by Frix last updated on 20/Aug/22

$$\frac{x^2-2x}{16-2x^3}=-\frac{x}{2(x^2+2x+4)}$$$$3^{\frac{x}{2(x^2+2x+4)}}=3^{\frac{1}{2x}}$$$$\frac{x}{2(x^2+2x+4)}=\frac{1}{2x}$$$$4x+8=0$$$$x=-2$$

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