Question Number 22220 by Tinkutara last updated on 13/Oct/17
$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{integral}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{4log}_{{x}/\mathrm{2}} \left(\sqrt{{x}}\right)+\mathrm{2log}_{\mathrm{4}{x}} \left({x}^{\mathrm{2}} \right)= \\ $$$$\mathrm{3log}_{\mathrm{2}{x}} \left({x}^{\mathrm{3}} \right)\:\mathrm{is} \\ $$
Answered by ajfour last updated on 13/Oct/17
![4log _(x/2) [(√2)((√(x/2)))]+2log _(4x) [(((4x)^2 )/(16))] =3log _(2x) [(((2x)^3 )/8)] ⇒ 4log _(x/2) (√2)+4((1/2))+2×2−2log _(4x) 16 =3×3−3log _(2x) 8 ((4log _2 (√2))/(log _2 x−1))−((2log _2 16)/(log _2 x+2))=3−((3log _2 8)/(log _2 x+1)) let log _2 x=t , then (2/(t−1))−(8/(t+2))=3−(9/(t+1)) if t≠1, −2, −1 then (2/(t−1))+(9/(t+1))=3+(8/(t+2)) ⇒ ((2t+2+9t−9)/(t^2 −1)) = ((3t+6+8)/(t+2)) (11t−7)(t+2)=(t^2 −1)(3t+14) 11t^2 +15t−14=3t^3 +14t^2 −3t−14 3t^3 +3t^2 −18t=0 ⇒ t(t^2 +t^2 −6)=0 or t(t+3)(t−2)=0 t=log _2 x= 0, 2, −3 ⇒ x= 1, 4, (1/8) hence two integral solutions.](https://www.tinkutara.com/question/Q22225.png)
$$\mathrm{4log}\:_{{x}/\mathrm{2}} \left[\sqrt{\mathrm{2}}\left(\sqrt{{x}/\mathrm{2}}\right)\right]+\mathrm{2log}\:_{\mathrm{4}{x}} \left[\frac{\left(\mathrm{4}{x}\right)^{\mathrm{2}} }{\mathrm{16}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3log}\:_{\mathrm{2}{x}} \left[\frac{\left(\mathrm{2}{x}\right)^{\mathrm{3}} }{\mathrm{8}}\right] \\ $$$$\Rightarrow\:\mathrm{4log}\:_{{x}/\mathrm{2}} \sqrt{\mathrm{2}}+\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{2}×\mathrm{2}−\mathrm{2log}\:_{\mathrm{4}{x}} \mathrm{16} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}×\mathrm{3}−\mathrm{3log}\:_{\mathrm{2}{x}} \mathrm{8} \\ $$$$\frac{\mathrm{4log}\:_{\mathrm{2}} \sqrt{\mathrm{2}}}{\mathrm{log}\:_{\mathrm{2}} {x}−\mathrm{1}}−\frac{\mathrm{2log}\:_{\mathrm{2}} \mathrm{16}}{\mathrm{log}\:_{\mathrm{2}} {x}+\mathrm{2}}=\mathrm{3}−\frac{\mathrm{3log}\:_{\mathrm{2}} \mathrm{8}}{\mathrm{log}\:_{\mathrm{2}} {x}+\mathrm{1}} \\ $$$${let}\:\mathrm{log}\:_{\mathrm{2}} {x}={t}\:,\:{then} \\ $$$$\frac{\mathrm{2}}{{t}−\mathrm{1}}−\frac{\mathrm{8}}{{t}+\mathrm{2}}=\mathrm{3}−\frac{\mathrm{9}}{{t}+\mathrm{1}} \\ $$$$\:{if}\:{t}\neq\mathrm{1},\:−\mathrm{2},\:−\mathrm{1}\:\:\:{then} \\ $$$$\frac{\mathrm{2}}{{t}−\mathrm{1}}+\frac{\mathrm{9}}{{t}+\mathrm{1}}=\mathrm{3}+\frac{\mathrm{8}}{{t}+\mathrm{2}}\:\:\:\:\Rightarrow \\ $$$$\frac{\mathrm{2}{t}+\mathrm{2}+\mathrm{9}{t}−\mathrm{9}}{{t}^{\mathrm{2}} −\mathrm{1}}\:=\:\frac{\mathrm{3}{t}+\mathrm{6}+\mathrm{8}}{{t}+\mathrm{2}} \\ $$$$\left(\mathrm{11}{t}−\mathrm{7}\right)\left({t}+\mathrm{2}\right)=\left({t}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{3}{t}+\mathrm{14}\right) \\ $$$$\mathrm{11}{t}^{\mathrm{2}} +\mathrm{15}{t}−\mathrm{14}=\mathrm{3}{t}^{\mathrm{3}} +\mathrm{14}{t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{14} \\ $$$$\mathrm{3}{t}^{\mathrm{3}} +\mathrm{3}{t}^{\mathrm{2}} −\mathrm{18}{t}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:{t}\left({t}^{\mathrm{2}} +{t}^{\mathrm{2}} −\mathrm{6}\right)=\mathrm{0} \\ $$$${or}\:\:\:{t}\left({t}+\mathrm{3}\right)\left({t}−\mathrm{2}\right)=\mathrm{0} \\ $$$${t}=\mathrm{log}\:_{\mathrm{2}} {x}=\:\mathrm{0},\:\mathrm{2},\:−\mathrm{3} \\ $$$$\Rightarrow\:\:\:\:\boldsymbol{{x}}=\:\mathrm{1},\:\mathrm{4},\:\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${hence}\:\boldsymbol{{two}}\:{integral}\:{solutions}. \\ $$
Commented by Tinkutara last updated on 14/Oct/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$