Logarithm question

Hello, Shapeshift!

Quote:

Given: . $\log_2\left(\log_8X\right) \:=\:\log_8\left(\log_2X\right)$

Find: . $(\log_2X)^2$

Let: $\log_8Y \:=\:p \quad\Rightarrow\quad 8^p \:=\:Y \quad\Rightarrow\quad \left(2^3\right)^p \:=\:Y \quad\Rightarrow\quad 2^{3p} \:=\:Y$

Take logs (base 2): . $\log_2\left(2^{3p}\right) \:=\:\log_2Y \quad\Rightarrow\quad 3p\log_22 \:=\:\log_2Y$

Since $\log_22 = 1$ , we have: . $3p \:=\:\log_2Y \quad\Rightarrow\quad p \:=\:\tfrac{1}{3}\log_2Y$

. . Hence: . $\log_8Y \:=\:\tfrac{1}{3}\log_2Y$

$\text{The equation: }\;\log_2\underbrace{\left(\log_8X\right)} \;\;=\;\;\underbrace{\log_8\left(\log_2X\right)}$
. . $\text{becomes: }\;\log_2\overbrace{\left(\tfrac{1}{3}\log_2X\righ t)} \;=\;\overbrace{\tfrac{1}{3}\log_2\left(\log_2X\ri ght)}$

Multiply by 3: . $3\log_2\left(\tfrac{1}{3}\log_2X\right) \;=\;\log_2\left(\log_2X\right) \quad\Rightarrow\quad \log_2\left(\tfrac{1}{3}\log_2X\right)^3 \;=\;\log_2\left(\log_2X\right)$

Exponentiate both sides: . $\left(\tfrac{1}{3}\log_2X\right)^3 \;=\;\log_2X \quad\Rightarrow\quad \tfrac{1}{27}\left(\log_2X\right)^3 \;=\;\log_2X$

Multiply by 27: . $\left(\log_2X\right)^3 \;=\;27\log_2X \quad\Rightarrow\quad \left(\log_2X\right)^3 - \log_2X\;=\;0$

Factor: . $\log_2X\bigg[(\log_2X)^2 - 27\bigg] \;=\;0$

We have two equations to solve . . .

. . $(1)\;\;\log_2X \:=\:0\quad\hdots$ ] .which is not allowed in the original equation

. . $[2]\;\;\left(\log_2X\right)^2 - 27 \:=\:0 \quad\Rightarrow\quad {\color{blue}\left(\log_2X\right)^2 \:=\:27}$