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Question-198489

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Question Number 198489 by sonukgindia last updated on 21/Oct/23
Answered by cortano12 last updated on 21/Oct/23
x^(16(log _5 x)^3 −68log _5 x ) = 5^(−16) let log _5 x = t x^(16 t^3 −68t) = 5^(−16) (16t^3 −68t)t = −16 4t^4 −17t^2 +4 = 0 (4t^2 −1)(t^2 −4)=0 { ((t=log _5 x = ±(1/2)⇒x= (√5) ,x=(1/( (√5))))),((t=log _5 x=±2⇒x=5^2 ; x=(1/(25)))) :}
$$\:\:\mathrm{x}^{\mathrm{16}\left(\mathrm{log}\:_{\mathrm{5}} \mathrm{x}\right)^{\mathrm{3}} −\mathrm{68log}\:_{\mathrm{5}} \mathrm{x}\:} =\:\mathrm{5}^{−\mathrm{16}} \\ $$$$\:\:\mathrm{let}\:\mathrm{log}\:_{\mathrm{5}} \mathrm{x}\:=\:\mathrm{t} \\ $$$$\:\:\mathrm{x}^{\mathrm{16}\:\mathrm{t}^{\mathrm{3}} −\mathrm{68t}} \:=\:\mathrm{5}^{−\mathrm{16}} \\ $$$$\:\left(\mathrm{16t}^{\mathrm{3}} −\mathrm{68t}\right)\mathrm{t}\:=\:−\mathrm{16}\: \\ $$$$\:\:\:\mathrm{4t}^{\mathrm{4}} −\mathrm{17t}^{\mathrm{2}} +\mathrm{4}\:=\:\mathrm{0} \\ $$$$\:\:\:\left(\mathrm{4t}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{2}} −\mathrm{4}\right)=\mathrm{0} \\ $$$$\:\:\:\:\begin{cases}{\mathrm{t}=\mathrm{log}\:_{\mathrm{5}} \mathrm{x}\:=\:\pm\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{x}=\:\sqrt{\mathrm{5}}\:,\mathrm{x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}}\\{\mathrm{t}=\mathrm{log}\:_{\mathrm{5}} \mathrm{x}=\pm\mathrm{2}\Rightarrow\mathrm{x}=\mathrm{5}^{\mathrm{2}} \:;\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{25}}}\end{cases} \\ $$$$\:\: \\ $$
Commented by Rasheed.Sindhi last updated on 21/Oct/23
x^(16 t^3 −68t) = 5^(−16) ⇒^(???) (16t^3 −68t)t = −16
$$\:\:\mathrm{x}^{\mathrm{16}\:\mathrm{t}^{\mathrm{3}} −\mathrm{68t}} \:=\:\mathrm{5}^{−\mathrm{16}} \overset{???} {\Rightarrow}\:\left(\mathrm{16t}^{\mathrm{3}} −\mathrm{68t}\right)\mathrm{t}\:=\:−\mathrm{16} \\ $$
Commented by cortano12 last updated on 21/Oct/23
log _5 (x^(16t^3 −68t) )= log _5 (5^(−16) ) (16t^3 −68t)t = −16
$$\:\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{x}^{\mathrm{16t}^{\mathrm{3}} −\mathrm{68t}} \right)=\:\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{5}^{−\mathrm{16}} \right) \\ $$$$\:\left(\mathrm{16t}^{\mathrm{3}} −\mathrm{68t}\right)\mathrm{t}\:=\:−\mathrm{16}\: \\ $$
Commented by Hridiana last updated on 21/Oct/23
3+((t/2)+1)^(t+1) =∫t+1dt
$$\mathrm{3}+\left(\frac{{t}}{\mathrm{2}}+\mathrm{1}\right)^{{t}+\mathrm{1}} =\int{t}+\mathrm{1}{dt} \\ $$
Commented by Frix last updated on 21/Oct/23
(4t^2 −1)(t^2 −4)=0
$$\left(\mathrm{4}{t}^{\mathrm{2}} −\mathrm{1}\right)\left({t}^{\mathrm{2}} −\mathrm{4}\right)=\mathrm{0} \\ $$
Answered by Frix last updated on 21/Oct/23
x=e^t e^(((16t^4 )/(ln^3 5))−((68t^2 )/(ln 5))) =5^(−16) ((16t^4 )/(ln^3 5))−((68t^2 )/(ln 5))=−16ln 5 t^4 −((17ln^2 5)/4)t^2 +ln^4 5 =0 t=±((ln 5)/2)∨t=±2ln 5 x=(1/(25))∨x=((√5)/5)∨x=(√5)∨x=25
$${x}=\mathrm{e}^{{t}} \\ $$$$\mathrm{e}^{\frac{\mathrm{16}{t}^{\mathrm{4}} }{\mathrm{ln}^{\mathrm{3}} \:\mathrm{5}}−\frac{\mathrm{68}{t}^{\mathrm{2}} }{\mathrm{ln}\:\mathrm{5}}} =\mathrm{5}^{−\mathrm{16}} \\ $$$$\frac{\mathrm{16}{t}^{\mathrm{4}} }{\mathrm{ln}^{\mathrm{3}} \:\mathrm{5}}−\frac{\mathrm{68}{t}^{\mathrm{2}} }{\mathrm{ln}\:\mathrm{5}}=−\mathrm{16ln}\:\mathrm{5} \\ $$$${t}^{\mathrm{4}} −\frac{\mathrm{17ln}^{\mathrm{2}} \:\mathrm{5}}{\mathrm{4}}{t}^{\mathrm{2}} +\mathrm{ln}^{\mathrm{4}} \:\mathrm{5}\:=\mathrm{0} \\ $$$$ \\ $$$${t}=\pm\frac{\mathrm{ln}\:\mathrm{5}}{\mathrm{2}}\vee{t}=\pm\mathrm{2ln}\:\mathrm{5} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{25}}\vee{x}=\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\vee{x}=\sqrt{\mathrm{5}}\vee{x}=\mathrm{25} \\ $$
Answered by HomeAlone last updated on 21/Oct/23
x^(128+x^(√2) +log _x (−x)+1)
$${x}^{\mathrm{128}+{x}^{\sqrt{\mathrm{2}}} +\mathrm{log}\:_{{x}} \left(−{x}\right)+\mathrm{1}} \\ $$

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