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Question Number 20195 by mondodotto@gmail.com last updated on 23/Aug/17

Answered by Tinkutara last updated on 23/Aug/17

$$\mathrm{Let}x=2^y$$$$2^{y(y+4)}=2^5$$$$y^2+4y-5=0$$$$(y+5)(y-1)=0$$$$y=1,-5$$$$x=2,2^{-5}$$$$\mathrm{Hence}2\mathrm{and}\frac{1}{32}\mathrm{are}\mathrm{solutions}.$$

Commented by mondodotto@gmail.com last updated on 23/Aug/17

$$\mathrm{thank}\mathrm{you}\mathrm{really}\mathrm{appriciate}\mathrm{this}!$$

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