If $x=(4096)^{7+4 \sqrt{3}}$, then which of the following equals $64$ ?

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We need to find which expression involving $x$ equals 64, where $x=(4096)^{7+4\sqrt{3}}$.

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Let's first simplify 4096:

$4096 = 2^{12}$ (since $2^{10} = 1024$, so $2^{12} = 1024 \times 4 = 4096$)

So our expression becomes:

$x = (2^{12})^{7+4\sqrt{3}} = 2^{12(7+4\sqrt{3})} = 2^{84+48\sqrt{3}}$

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Express 64 as a power of 2:

$64 = 2^6$

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We need to find some value $a$ such that $x^a = 64$.

Since $x = 2^{84+48\sqrt{3}}$ and $64 = 2^6$:

$x^a = (2^{84+48\sqrt{3}})^a = 2^{a(84+48\sqrt{3})} = 2^6$

For this equation to hold:

$a(84+48\sqrt{3}) = 6$

Therefore: $a = \dfrac{6}{84+48\sqrt{3}}$

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To rationalize $\dfrac{6}{84+48\sqrt{3}}$, we use the conjugate $(84-48\sqrt{3})$:

$a = \dfrac{6}{84+48\sqrt{3}} \times \dfrac{84-48\sqrt{3}}{84-48\sqrt{3}} = \dfrac{6(84-48\sqrt{3})}{(84)^2-(48\sqrt{3})^2}$

Calculate the denominator:

• $(84)^2 = 7056$
• $(48\sqrt{3})^2 = 48^2 \times 3 = 2304 \times 3 = 6912$
• $(84)^2-(48\sqrt{3})^2 = 7056 - 6912 = 144$

So: $a = \dfrac{6(84-48\sqrt{3})}{144} = \dfrac{84-48\sqrt{3}}{24}$

Simplifying: $a = \dfrac{84}{24} - \dfrac{48\sqrt{3}}{24} = \dfrac{7}{2} - 2\sqrt{3}$

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Therefore: $x^{\dfrac{7}{2} - 2\sqrt{3}} = 64$

Using the property $a^{m-n} = \dfrac{a^m}{a^n}$:

$\dfrac{x^{\dfrac{7}{2}}}{x^{2\sqrt{3}}} = 64$

When dealing with expressions involving surds in exponents, rationalization is often needed. The conjugate method helps eliminate square roots from denominators, and powers can be separated using the rule $a^{m-n} = \dfrac{a^m}{a^n}$.