Question

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

• A. \(\frac{x^7}{x^{2\sqrt3}}\)
• B. \(\frac{x^7}{x^{4\sqrt3}}\)
• C. \(\frac{x\frac{7}{2}}{x^{\frac{4}{\sqrt3}}}\)
• D. \(\frac{x\frac{7}{2}}{x^{2\sqrt3}}\)

Answer 1

The Correct Option is D

Let \( x = (4096)^{7 + 4\sqrt{3}} \).

We know that \( 4096 = 2^{12} \), so:

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

Now we evaluate each expression to see which one equals \( 64 = 2^6 \).

Option 1: \( \frac{x^7}{x^{2\sqrt{3}}} = x^{7 - 2\sqrt{3}} \)

\[ x^{7 - 2\sqrt{3}} = \left(2^{84 + 48\sqrt{3}}\right)^{7 - 2\sqrt{3}} = 2^{(84 + 48\sqrt{3})(7 - 2\sqrt{3})} \]

Expanding the exponent: \[ 84 \cdot 7 + 84 \cdot (-2\sqrt{3}) + 48\sqrt{3} \cdot 7 + 48\sqrt{3} \cdot (-2\sqrt{3}) \] \[ = 588 - 168\sqrt{3} + 336\sqrt{3} - 288 = 300 + 168\sqrt{3} \]

So the result is: \[ 2^{300 + 168\sqrt{3}} \neq 64 \]

Option 2: \( \frac{x^7}{x^{4\sqrt{3}}} = x^{7 - 4\sqrt{3}} \)

\[ x^{7 - 4\sqrt{3}} = \left(2^{84 + 48\sqrt{3}}\right)^{7 - 4\sqrt{3}} = 2^{(84 + 48\sqrt{3})(7 - 4\sqrt{3})} \]

Expanding the exponent: \[ 84 \cdot 7 + 84 \cdot (-4\sqrt{3}) + 48\sqrt{3} \cdot 7 + 48\sqrt{3} \cdot (-4\sqrt{3}) \] \[ = 588 - 336\sqrt{3} + 336\sqrt{3} - 576 = 12 \]

So the result is: \[ 2^{12} = 4096 \neq 64 \]

Option 3: \( \frac{x^{7/2}}{x^{4/\sqrt{3}}} = x^{\frac{7}{2} - \frac{4}{\sqrt{3}}} \)

\[ x^{\frac{7}{2} - \frac{4}{\sqrt{3}}} = \left(2^{84 + 48\sqrt{3}}\right)^{\frac{7}{2} - \frac{4}{\sqrt{3}}} \]

This does not simplify neatly and will not give an integer power of 2. So this is also incorrect.

Option 4: \( \frac{x^{7/2}}{x^{2\sqrt{3}}} = x^{\frac{7}{2} - 2\sqrt{3}} \)

\[ x^{\frac{7}{2} - 2\sqrt{3}} = \left(2^{84 + 48\sqrt{3}}\right)^{\frac{7}{2} - 2\sqrt{3}} = 2^{(84 + 48\sqrt{3})(\frac{7}{2} - 2\sqrt{3})} \]

Expanding the exponent: \[ 84 \cdot \frac{7}{2} + 84 \cdot (-2\sqrt{3}) + 48\sqrt{3} \cdot \frac{7}{2} + 48\sqrt{3} \cdot (-2\sqrt{3}) \] \[ = 294 - 168\sqrt{3} + 168\sqrt{3} - 288 = 6 \]

So the result is: \[ 2^6 = 64 \]

Final Answer: \( \boxed{\frac{x^{7/2}}{x^{2\sqrt{3}}}} \)

Answer 2

Given:

\( x = (4096)^{7 + 4\sqrt{3}} \)

Step 1: Convert the base

\( 4096 = 2^{12} \), so:

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

Step 2: Compute \( x^{\frac{7}{2}} \)

\[ x^{\frac{7}{2}} = \left(2^{12(7 + 4\sqrt{3})}\right)^{\frac{7}{2}} = 2^{\frac{7}{2} \cdot 12(7 + 4\sqrt{3})} = 2^{42(7 + 4\sqrt{3})} \]

Step 3: Compute \( x^{2\sqrt{3}} \)

\[ x^{2\sqrt{3}} = \left(2^{12(7 + 4\sqrt{3})}\right)^{2\sqrt{3}} = 2^{24\sqrt{3}(7 + 4\sqrt{3})} \]

Step 4: Evaluate the required expression

\[ \frac{x^{\frac{7}{2}}}{x^{2\sqrt{3}}} = \frac{2^{42(7 + 4\sqrt{3})}}{2^{24\sqrt{3}(7 + 4\sqrt{3})}} = 2^{(7 + 4\sqrt{3}) \cdot (42 - 24\sqrt{3})} \]

Step 5: Use identity \( (a + b)(a - b) = a^2 - b^2 \)

\[ (7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 49 - 48 = 1 \]

\[ \Rightarrow \frac{x^{\frac{7}{2}}}{x^{2\sqrt{3}}} = 2^{6} = 64 \]

 Final Answer:

\[ \frac{x^{\frac{7}{2}}}{x^{2\sqrt{3}}} = 64 \]

Correct Option: (D)