Math Reasonings Solved Exercises

Exercise 4

If:

\[ x = 4^{ \frac{k}{3} + 2L } \left( \sqrt[3]{4} \right)^{ - (K + 4L) } \left( 16^{ 1 - \frac{L}{3} } \right), \]

then \(x\) equals to:

If   \( x = 4^{ \frac{k}{3} + 2L } \left( \sqrt[3]{4} \right)^{ - (K + 4L) } \left( 16^{ 1 - \frac{L}{3} } \right) \),   then \(x\) equals to:

  1. 8

  2. 4096

  3. 1

  4. 64

  5. 16

Try to solve it before checking the answer.

  1. 16

Considering that \( \sqrt[3]{4} = 4^{\frac{1}{3}} \)   y   \(16 = 4^2\), we have:
\[ \begin{aligned} x &= 4^{ \frac{k}{3} + 2L } \left( \sqrt[3]{4} \right)^{ - (K + 4L) } \left( 16^{ 1 - \frac{L}{3} } \right) \\[.5em] &= 4^{ \frac{k}{3} + 2L } \left( 4^{\frac{1}{3}} \right)^{ - (k + 4L) } \left( \left( 4^2 \right)^{ 1 - \frac{L}{3} } \right) \\[.5em] &= 4^{ \frac{k}{3} + 2L } \times 4^{ \frac{ -k + 4L }{3} } \times 4^{ 2 \left( 1 - \frac{L}{3} \right) } \\[.5em] &= 4^{ \frac{k}{3} + 2L } \times 4^{ \frac{-k - 4L}{3} } \times 4^{ 2 - \frac{2L}{3} } \\[.5em] &= 4^{ \frac{k}{3} + 2L + \frac{ -k - 4L }{3} + 2 - \frac{2L}{3} } \\[.5em] &= 4^{ \frac{ k + 6L - k - 4L + 6 - 2L }{3} } \\[.5em] &= 4^{\frac{6}{3}} \\[.5em] &= 4^2 \\[.5em] &= \boldsymbol{16} \end{aligned} \]
\[ \begin{aligned} x &= 4^{ \frac{k}{3} + 2L } \left( \sqrt[3]{4} \right)^{ - (K + 4L) } \left( 16^{ 1 - \frac{L}{3} } \right) \\[.5em] &= 4^{ \frac{k}{3} + 2L } \left( 4^{\frac{1}{3}} \right)^{ - (k + 4L) } \left( \left( 4^2 \right)^{ 1 - \frac{L}{3} } \right) \\[.5em] &= 4^{ \frac{k}{3} + 2L } \times 4^{ \frac{ -k + 4L }{3} } \times 4^{ 2 \left( 1 - \frac{L}{3} \right) } \\[.5em] &= 4^{ \frac{k}{3} + 2L } \times 4^{ \frac{-k - 4L}{3} } \times 4^{ 2 - \frac{2L}{3} } \\[.5em] &= 4^{ \frac{k}{3} + 2L + \frac{ -k - 4L }{3} + 2 - \frac{2L}{3} } \\[.5em] &= 4^{ \frac{ k + 6L - k - 4L + 6 - 2L }{3} } \\[.5em] &= 4^{\frac{6}{3}} \\[.5em] &= 4^2 \\[.5em] &= \boldsymbol{16} \end{aligned} \]

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