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Question Number 167821 by mathlove last updated on 26/Mar/22

Commented by dangduomg last updated on 26/Mar/22

⇒ E = (4^(1/5) )^((5^(1/4) )^E ) = 4^(1/5×(5^(1/4) )^E ) = 4^(1/5×5^(E/4) ) = 4^5^(E/4−1)

$$\Rightarrow\:{E}\:=\:\left(\mathrm{4}^{\mathrm{1}/\mathrm{5}} \right)^{\left(\mathrm{5}^{\mathrm{1}/\mathrm{4}} \right)^{{E}} } \\ $$$$=\:\mathrm{4}^{\mathrm{1}/\mathrm{5}×\left(\mathrm{5}^{\mathrm{1}/\mathrm{4}} \right)^{{E}} } \\ $$$$=\:\mathrm{4}^{\mathrm{1}/\mathrm{5}×\mathrm{5}^{{E}/\mathrm{4}} } \\ $$$$=\:\mathrm{4}^{\mathrm{5}^{{E}/\mathrm{4}−\mathrm{1}} } \\ $$

Commented by aleks041103 last updated on 27/Mar/22

E=4: 4^5^(4/4−1) =4^5^0 =4^1 =4=E ⇒E=4 is a soln. We should note that it is not the only solution log_4 E=5^(E/4−1) x=5log_4 E ⇒x=5.5^((1/4)4^(x/5) −1) =5^4^(x/5−1) =x ⇒log_5 x=4^(x/5−1) y=4log_5 x=4log_5 (5log_4 E) ⇒y=4.4^(5^(y/4) /5−1) =4^5^(y/4−1) E=4+4log_5 (log_4 E) let f(E)=E−4−4log_5 (log_4 E) from E=4^5^(E/4−1) we see that E>1 f(1)→+∞>0 f(16)=12−4log_5 (2) log_5 (2)<1⇒f(16)>12−4=8>0 f(16)>0 BUT f(2)=−2−4log_5 (log_4 2)= =−2−4log_5 (1/2)= =4log_5 (2)−2= =log_5 (2^4 )−log_5 (5^2 )= =log_5 (((16)/(25)))=log_5 (smth<1)<0 ⇒f(0)>0 (1) f(2)<0 (2) f(16)>0 (3) From (1) and (2) we see that there is at least 1 soln. between 0 and 2 From (2) and (3) we see that there is at least 1 soln. between 2 and 16( we know that one of those is 4) By using Desmos we find there are only 2 solns.: E=4 and E≈1,753

$${E}=\mathrm{4}: \\ $$$$\mathrm{4}^{\mathrm{5}^{\mathrm{4}/\mathrm{4}−\mathrm{1}} } =\mathrm{4}^{\mathrm{5}^{\mathrm{0}} } =\mathrm{4}^{\mathrm{1}} =\mathrm{4}={E} \\ $$$$\Rightarrow{E}=\mathrm{4}\:{is}\:{a}\:{soln}. \\ $$$${We}\:{should}\:{note}\:{that}\:{it}\:{is}\:{not}\:{the}\:{only} \\ $$$${solution} \\ $$$${log}_{\mathrm{4}} {E}=\mathrm{5}^{{E}/\mathrm{4}−\mathrm{1}} \\ $$$${x}=\mathrm{5}{log}_{\mathrm{4}} {E} \\ $$$$\Rightarrow{x}=\mathrm{5}.\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}\mathrm{4}^{{x}/\mathrm{5}} −\mathrm{1}} =\mathrm{5}^{\mathrm{4}^{{x}/\mathrm{5}−\mathrm{1}} } ={x} \\ $$$$\Rightarrow{log}_{\mathrm{5}} {x}=\mathrm{4}^{{x}/\mathrm{5}−\mathrm{1}} \\ $$$${y}=\mathrm{4}{log}_{\mathrm{5}} {x}=\mathrm{4}{log}_{\mathrm{5}} \left(\mathrm{5}{log}_{\mathrm{4}} {E}\right) \\ $$$$\Rightarrow{y}=\mathrm{4}.\mathrm{4}^{\mathrm{5}^{{y}/\mathrm{4}} /\mathrm{5}−\mathrm{1}} =\mathrm{4}^{\mathrm{5}^{{y}/\mathrm{4}−\mathrm{1}} } \\ $$$${E}=\mathrm{4}+\mathrm{4}{log}_{\mathrm{5}} \left({log}_{\mathrm{4}} {E}\right) \\ $$$${let}\:{f}\left({E}\right)={E}−\mathrm{4}−\mathrm{4}{log}_{\mathrm{5}} \left({log}_{\mathrm{4}} {E}\right) \\ $$$${from}\:{E}=\mathrm{4}^{\mathrm{5}^{{E}/\mathrm{4}−\mathrm{1}} } \:{we}\:{see}\:{that}\:{E}>\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)\rightarrow+\infty>\mathrm{0} \\ $$$${f}\left(\mathrm{16}\right)=\mathrm{12}−\mathrm{4}{log}_{\mathrm{5}} \left(\mathrm{2}\right) \\ $$$${log}_{\mathrm{5}} \left(\mathrm{2}\right)<\mathrm{1}\Rightarrow{f}\left(\mathrm{16}\right)>\mathrm{12}−\mathrm{4}=\mathrm{8}>\mathrm{0} \\ $$$${f}\left(\mathrm{16}\right)>\mathrm{0} \\ $$$${BUT} \\ $$$${f}\left(\mathrm{2}\right)=−\mathrm{2}−\mathrm{4}{log}_{\mathrm{5}} \left({log}_{\mathrm{4}} \mathrm{2}\right)= \\ $$$$=−\mathrm{2}−\mathrm{4}{log}_{\mathrm{5}} \left(\mathrm{1}/\mathrm{2}\right)= \\ $$$$=\mathrm{4}{log}_{\mathrm{5}} \left(\mathrm{2}\right)−\mathrm{2}= \\ $$$$={log}_{\mathrm{5}} \left(\mathrm{2}^{\mathrm{4}} \right)−{log}_{\mathrm{5}} \left(\mathrm{5}^{\mathrm{2}} \right)= \\ $$$$={log}_{\mathrm{5}} \left(\frac{\mathrm{16}}{\mathrm{25}}\right)={log}_{\mathrm{5}} \left({smth}<\mathrm{1}\right)<\mathrm{0} \\ $$$$\Rightarrow{f}\left(\mathrm{0}\right)>\mathrm{0}\:\left(\mathrm{1}\right) \\ $$$${f}\left(\mathrm{2}\right)<\mathrm{0}\:\left(\mathrm{2}\right) \\ $$$${f}\left(\mathrm{16}\right)>\mathrm{0}\:\left(\mathrm{3}\right) \\ $$$${From}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right)\:{we}\:{see}\:{that}\:{there}\:{is} \\ $$$${at}\:{least}\:\mathrm{1}\:{soln}.\:{between}\:\mathrm{0}\:{and}\:\mathrm{2} \\ $$$${From}\:\left(\mathrm{2}\right)\:{and}\:\left(\mathrm{3}\right)\:{we}\:{see}\:{that}\:{there}\:{is} \\ $$$${at}\:{least}\:\mathrm{1}\:{soln}.\:{between}\:\mathrm{2}\:{and}\:\mathrm{16}\left(\:{we}\:{know}\right. \\ $$$$\left.{that}\:{one}\:{of}\:{those}\:{is}\:\mathrm{4}\right) \\ $$$${By}\:{using}\:{Desmos}\:{we}\:{find}\:{there}\:{are} \\ $$$${only}\:\mathrm{2}\:{solns}.:\:{E}=\mathrm{4}\:{and}\:{E}\approx\mathrm{1},\mathrm{753} \\ $$

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