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2-4-y-2-2-y-y-2-8-please-help-me-find-y-pls-




Question Number 17759 by chux last updated on 10/Jul/17
2^(4(y^2 −2)) =y^((y^2 −8))       please help me find y... pls!
$$\mathrm{2}^{\mathrm{4}\left(\mathrm{y}^{\mathrm{2}} −\mathrm{2}\right)} =\mathrm{y}^{\left(\mathrm{y}^{\mathrm{2}} −\mathrm{8}\right)} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{find}\:\mathrm{y}…\:\mathrm{pls}! \\ $$
Answered by mrW1 last updated on 10/Jul/17
no analytic solution possible.  y≈1.23066
$$\mathrm{no}\:\mathrm{analytic}\:\mathrm{solution}\:\mathrm{possible}. \\ $$$$\mathrm{y}\approx\mathrm{1}.\mathrm{23066} \\ $$
Commented by chux last updated on 10/Jul/17
Commented by chux last updated on 10/Jul/17
when trying to understand your  solution to this question i arrived  at this equation...... This means   that i did not really understand  your approach majorly at the  end of the solution. I ll love it if  you shed more light to it sir.    Thanks for always assisting.
$$\mathrm{when}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{your} \\ $$$$\mathrm{solution}\:\mathrm{to}\:\mathrm{this}\:\mathrm{question}\:\mathrm{i}\:\mathrm{arrived} \\ $$$$\mathrm{at}\:\mathrm{this}\:\mathrm{equation}……\:\mathrm{This}\:\mathrm{means}\: \\ $$$$\mathrm{that}\:\mathrm{i}\:\mathrm{did}\:\mathrm{not}\:\mathrm{really}\:\mathrm{understand} \\ $$$$\mathrm{your}\:\mathrm{approach}\:\mathrm{majorly}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{end}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solution}.\:\mathrm{I}\:\mathrm{ll}\:\mathrm{love}\:\mathrm{it}\:\mathrm{if} \\ $$$$\mathrm{you}\:\mathrm{shed}\:\mathrm{more}\:\mathrm{light}\:\mathrm{to}\:\mathrm{it}\:\mathrm{sir}. \\ $$$$ \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{always}\:\mathrm{assisting}. \\ $$
Commented by mrW1 last updated on 10/Jul/17
how can i help you sir?  there is no general methods for your  equations.
$$\mathrm{how}\:\mathrm{can}\:\mathrm{i}\:\mathrm{help}\:\mathrm{you}\:\mathrm{sir}? \\ $$$$\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{general}\:\mathrm{methods}\:\mathrm{for}\:\mathrm{your} \\ $$$$\mathrm{equations}. \\ $$
Commented by chux last updated on 10/Jul/17
i actually understand your step  from the top until the last step  before you got x=1    x^(2(x^2 −1/x^2 )) =2^(2(1−1/x^2 ))     x=1 and 1−(1/x^2 )=0    i dont know how you got the result  x=1
$$\mathrm{i}\:\mathrm{actually}\:\mathrm{understand}\:\mathrm{your}\:\mathrm{step} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{top}\:\mathrm{until}\:\mathrm{the}\:\mathrm{last}\:\mathrm{step} \\ $$$$\mathrm{before}\:\mathrm{you}\:\mathrm{got}\:\mathrm{x}=\mathrm{1} \\ $$$$ \\ $$$$\mathrm{x}^{\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}/\mathrm{x}^{\mathrm{2}} \right)} =\mathrm{2}^{\mathrm{2}\left(\mathrm{1}−\mathrm{1}/\mathrm{x}^{\mathrm{2}} \right)} \\ $$$$ \\ $$$$\mathrm{x}=\mathrm{1}\:\mathrm{and}\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{i}\:\mathrm{dont}\:\mathrm{know}\:\mathrm{how}\:\mathrm{you}\:\mathrm{got}\:\mathrm{the}\:\mathrm{result} \\ $$$$\mathrm{x}=\mathrm{1} \\ $$
Commented by mrW1 last updated on 10/Jul/17
1^p =2^0 =1  as i haved said there is no general  method one can follow blindly. one  has to try and see if error.
$$\mathrm{1}^{\mathrm{p}} =\mathrm{2}^{\mathrm{0}} =\mathrm{1} \\ $$$$\mathrm{as}\:\mathrm{i}\:\mathrm{haved}\:\mathrm{said}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{general} \\ $$$$\mathrm{method}\:\mathrm{one}\:\mathrm{can}\:\mathrm{follow}\:\mathrm{blindly}.\:\mathrm{one} \\ $$$$\mathrm{has}\:\mathrm{to}\:\mathrm{try}\:\mathrm{and}\:\mathrm{see}\:\mathrm{if}\:\mathrm{error}. \\ $$
Commented by chux last updated on 10/Jul/17
thanks sir
$$\mathrm{thanks}\:\mathrm{sir} \\ $$

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