Question #53128

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Question #53128

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Answer 1 · 2017-03-28T09:02:29

$y \approx 1.4434035067785973 `$ $y approx 1.4434035067785973`$

Explanation:

${(4 y)}^{y} = \frac{{(8^{5})}^{y}}{8^{6}}$ $(4y)^y=((8^5)^y)/8^6$

or

${(\frac{4 y}{8^{5}})}^{y} = 8^{- 6}$ $((4y)/(8^5))^y=8^(-6)$

Now calling $z = \frac{4 y}{8^{5}}$ $z=(4y)/(8^5)$ we have

$z^{\frac{8^{5} z}{4}} = z^{2 \times 8^{4} z} = 8^{- 6}$ $z^((8^5 z)/4) = z^(2xx 8^4 z) =8^(-6)$ or

$z^{z} = {(8^{- 6})}^{\frac{1}{2 \times 8^{4}}} = {(8^{- 3})}^{\frac{1}{8^{4}}}$ $z^z=(8^(-6))^(1/(2 xx 8^4)) = (8^(-3))^(1/8^4)$

Now the equation

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

$z = \frac{{log}_{e} a}{W ({log}_{e} a)}$ $z=log_ea/(W(log_ea))$ where $W (\cdot)$ $W(cdot)$ is the Lambert function

obtaining $z = - \frac{9 {log}_{e} 2}{4096 W (- \frac{9 {log}_{e} 2}{4096})} \approx 0.00017619671713605924$ $z = -(9 Log_e2)/(4096 W(-(9 Log_e2)/4096)) approx 0.00017619671713605924$ and $y \approx 1.4434035067785973$ $y approx 1.4434035067785973$

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Question #53128

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