# Question 53128

Mar 28, 2017

y approx 1.4434035067785973`#

#### Explanation:

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

or

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

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

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

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

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

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

obtaining $z = - \frac{9 L o {g}_{e} 2}{4096 W \left(- \frac{9 L o {g}_{e} 2}{4096}\right)} \approx 0.00017619671713605924$ and $y \approx 1.4434035067785973$