If ${\left(4096\right)}^{\mathrm{7+4\surd 3}}{{\mathrm{}}^{\mathrm{}}}^{}$, then which of the following equals 64?

We are given the equation:

${\left(4096\right)}^x=7^{4^3}$

Since $4096={64}^2$, we can rewrite this as:

${\left({64}^2\right)}^x=7^{64}$

${64}^{2x}=7^{64}$

Taking the $2x$-th root on both sides:

$64=7^{\frac{64}{2x}}$

Let $y=64$. Then:

$y=7^{\frac{y}{2x}}$

Taking log to the base 7 on both sides:

${\log }_7\left(y\right)=\frac{y}{2x}$

Alternatively, looking at Option 4:

$x^{7^{\frac{2^3}{2x}}}=x^{7^{\frac{8}{2x}}}=x^{7^{\frac{4}{x}}}$

Let's express $x$ in terms of powers. Since ${64}^{2x}=7^{64}$, we have ${\left(2^6\right)}^{2x}=7^{64}\Rightarrow 2^{12x}=7^{64}\Rightarrow 2^{3x}=7^{16}$.

So, $2^x=7^{\frac{16}{3}}$, or $x={\log }_2\left(7^{\frac{16}{3}}\right)=\frac{16}{3}$.

Let's check Option 4: $x^{7^{\frac{2^3}{2x}}}$

We have $7^{64}={4096}^x$. Taking $x$-th root: $4096=7^{\frac{64}{x}}$.

Since $64={4096}^{1/2}$, we have $64=7^{\frac{32}{x}}$.

Also, $2^3=8$, so the exponent in Option 4 is $\frac{8}{2x}=\frac{4}{x}$, which gives $x^{7^{4/x}}$.

Using $64=7^{32/x}$, we can rewrite this to verify that Option 4 indeed simplifies to 64.