Question #53128

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Answer 1 · 2017-03-28T09:02:29

#y approx 1.4434035067785973`#

#(4y)^y=((8^5)^y)/8^6#

or

#((4y)/(8^5))^y=8^(-6)#

Now calling #z=(4y)/(8^5)# we have

#z^((8^5 z)/4) = z^(2xx 8^4 z) =8^(-6)# or

#z^z=(8^(-6))^(1/(2 xx 8^4)) = (8^(-3))^(1/8^4)#

Now the equation

#z^z= a# is solved with the help of the so called Lambert function
https://en.wikipedia.org/wiki/Lambert_W_function which states

#z=log_ea/(W(log_ea))# where #W(cdot)# is the Lambert function

obtaining #z = -(9 Log_e2)/(4096 W(-(9 Log_e2)/4096)) approx 0.00017619671713605924# and #y approx 1.4434035067785973#