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

Commented by dangduomg last updated on 26/Mar/22

$$\Rightarrow E={\left(4^{1/5}\right)}^{{\left(5^{1/4}\right)}^E}$$$$=4^{1/5\times {\left(5^{1/4}\right)}^E}$$$$=4^{1/5\times 5^{E/4}}$$$$=4^{5^{E/4-1}}$$

Commented by aleks041103 last updated on 27/Mar/22

$$E=4:$$$$4^{5^{4/4-1}}=4^{5^0}=4^1=4=E$$$$\Rightarrow E=4isasoln.$$$$Weshouldnotethatitisnottheonly$$$$solution$$$${log}_{4}E=5^{E/4-1}$$$$x=5{log}_{4}E$$$$\Rightarrow x=5.5^{\frac{1}{4}{4}^{x/5}-1}=5^{4^{x/5-1}}=x$$$$\Rightarrow {log}_{5}x=4^{x/5-1}$$$$y=4{log}_{5}x=4{log}_5\left(5{log}_{4}E\right)$$$$\Rightarrow y=4.4^{5^{y/4}/5-1}=4^{5^{y/4-1}}$$$$E=4+4{log}_5\left({log}_{4}E\right)$$$$letf\left(E\right)=E-4-4{log}_5\left({log}_{4}E\right)$$$$fromE=4^{5^{E/4-1}}weseethatE>1$$$$f\left(1\right)\to +\mathrm{\infty }>0$$$$f\left(16\right)=12-4{log}_5\left(2\right)$$$${log}_5\left(2\right)<1\Rightarrow f\left(16\right)>12-4=8>0$$$$f\left(16\right)>0$$$$BUT$$$$f\left(2\right)=-2-4{log}_5\left({log}_{4}2\right)=$$$$=-2-4{log}_5\left(1/2\right)=$$$$=4{log}_5\left(2\right)-2=$$$$={log}_5\left(2^4\right)-{log}_5\left(5^2\right)=$$$$={log}_5\left(\frac{16}{25}\right)={log}_5(smth<1)<0$$$$\Rightarrow f\left(0\right)>0\left(1\right)$$$$f\left(2\right)<0\left(2\right)$$$$f\left(16\right)>0\left(3\right)$$$$From\left(1\right)and\left(2\right)weseethatthereis$$$$atleast1soln.between0and2$$$$From\left(2\right)and\left(3\right)weseethatthereis$$$$atleast1soln.between2and16(weknow$$$$thatoneofthoseis4)$$$$ByusingDesmoswefindthereare$$$$only2solns.:E=4andE\approx 1,753$$

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