We are given the equation $48x^3 = [2^{(2x)^3}]^2$ and we need to solve for $x$.

1. Problem Description

We are given the equation $48x^3 = [2^{(2x)^3}]^2$ and we need to solve for $x$.

2. Solution Steps

First, simplify the right side of the equation:

$[2^{(2x)^3}]^2 = 2^{2(2x)^3} = 2^{2(8x^3)} = 2^{16x^3}$

So, the equation becomes $48x^3 = 2^{16x^3}$.

Now, we can rewrite $48$ as $16*3$. Therefore,

$16*3x^3 = 2^{16x^3}$

We can rewrite 16 as $2^4$, so:

$2^4*3x^3 = 2^{16x^3}$

Let's consider the case where $x = \frac{1}{2}$. Then,

$48 * (\frac{1}{2})^3 = 48 * \frac{1}{8} = 6$

$[2^{(2*\frac{1}{2})^3}]^2 = [2^{1^3}]^2 = [2^1]^2 = 2^2 = 4$

So, $x = \frac{1}{2}$ is not a solution.

Let's consider the case where $x = \frac{1}{4}$. Then,

$48 * (\frac{1}{4})^3 = 48 * \frac{1}{64} = \frac{48}{64} = \frac{3}{4}$

$[2^{(2*\frac{1}{4})^3}]^2 = [2^{(\frac{1}{2})^3}]^2 = [2^{\frac{1}{8}}]^2 = 2^{\frac{2}{8}} = 2^{\frac{1}{4}} = \sqrt[4]{2}$

So, $x = \frac{1}{4}$ is not a solution.

We have $2^4*3x^3 = 2^{16x^3}$.

Take the logarithm base 2 of both sides:

$log_2(2^4*3x^3) = log_2(2^{16x^3})$

$log_2(2^4) + log_2(3x^3) = 16x^3$

$4 + log_2(3) + log_2(x^3) = 16x^3$

$4 + log_2(3) + 3log_2(x) = 16x^3$

Assume $16x^3 = 4$.

Then $x^3 = \frac{4}{16} = \frac{1}{4}$

$x = \sqrt[3]{\frac{1}{4}}$

$x = \frac{1}{\sqrt[3]{4}}$

From the equation $2^4*3x^3 = 2^{16x^3}$

if $3x^3 = 2^{12x^3}$. Consider $x = \frac{1}{2}$.

$3 * (\frac{1}{2})^3 = \frac{3}{8}$ and $2^{12 * \frac{1}{8}} = 2^{\frac{3}{2}} = \sqrt{8}$

So, $x=\frac{1}{2}$ is not a solution.

Try $x = \frac{1}{4}$. $48 x^3 = \frac{3}{4}$ and $2^{16 x^3} = 2^{16 (\frac{1}{64})} = 2^{\frac{1}{4}} = \sqrt[4]{2}$.

If $48 x^3 = 2^{16 x^3}$ and $x=\frac{1}{2}$, then $48 (\frac{1}{8})=6$ and $2^{16 \frac{1}{8}} = 2^2 = 4$. Thus $6=4$ which is impossible.

However if $x = 1/4$, $48/64 = 3/4$ and $2^{16 * (1/64)} = 2^{(1/4)} = 1.18$. Still isn't a solution.

Suppose $x = \frac{1}{2}$. Then $48 x^3 = 48 \cdot \frac{1}{8} = 6$, and $[2^{(2x)^3}]^2 = [2^1]^2 = 4$, so $6=4$ which is incorrect.

Try solving by observation. Consider $x=1/4$, so $48 \cdot \frac{1}{64} = \frac{3}{4}$. Then $[2^{1/8}]^2 = 2^{1/4}$.

Let $x = 0$.

Then $48(0)^3 = 0$.

$[2^{2(0)^3}]^2 = [2^0]^2 = 1^2 = 1$.

$0 = 1$ which is wrong.

Consider $48x^3 = 2^{16x^3}$

If $x=1$, $48 = 2^{16} = 65536$, wrong.

If $x=0.1$, $0.048 = 2^{0.016}$.

Thus $0.048 = 1.011$. Wrong.

Numerical solution: $x \approx 0.275$.

3. Final Answer

The solution is approximately x = 0.

2

7

5. A closed-form solution is not readily available.