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Question-20195




Question Number 20195 by mondodotto@gmail.com last updated on 23/Aug/17
Answered by Tinkutara last updated on 23/Aug/17
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)   Hence 2 and (1/(32)) are solutions.
$$\mathrm{Let}\:{x}\:=\:\mathrm{2}^{{y}} \\ $$$$\mathrm{2}^{{y}\left({y}+\mathrm{4}\right)} \:=\:\mathrm{2}^{\mathrm{5}} \\ $$$${y}^{\mathrm{2}} \:+\:\mathrm{4}{y}\:−\:\mathrm{5}\:=\:\mathrm{0} \\ $$$$\left({y}\:+\:\mathrm{5}\right)\left({y}\:−\:\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$${y}\:=\:\mathrm{1},\:−\mathrm{5} \\ $$$${x}\:=\:\mathrm{2},\:\mathrm{2}^{−\mathrm{5}} \\ $$$$\mathrm{Hence}\:\mathrm{2}\:\mathrm{and}\:\frac{\mathrm{1}}{\mathrm{32}}\:\mathrm{are}\:\mathrm{solutions}. \\ $$
Commented by mondodotto@gmail.com last updated on 23/Aug/17
thank you really appriciate this!
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{really}\:\mathrm{appriciate}\:\mathrm{this}! \\ $$

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