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 $y approx 1.4434035067785973$