\( \left. \begin{array} { l } { 16 ^ { 3 } \cdot 3 ^ { 3 } = x ^ { 3 } ; x = } \\ { 4 ^ { 2 } \cdot x ^ { 2 } \cdot 3 ^ { 2 } = 84 ^ { 2 } ; x = } \\ { 3 ^ { 2 } \cdot 3 ^ { 5 } \cdot 3 ^ { x } = 3 ^ { 11 } ; x = } \\ { ( x ^ { 2 } ) ^ { 5 } = 1024 ; x = } \end{array} \right. \)
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Bonus Knowledge
Let's dive into these equations and solve them step by step! 1. **For the first equation:** \[ 16^{3} \cdot 3^{3} = x^{3} \] We can rewrite \( 16 \) as \( 4^{2} \), so \( 16^{3} = (4^{2})^{3} = 4^{6} \). Thus, we have: \[ 4^{6} \cdot 3^{3} = x^{3} \] Taking the cube root gives \( x = (4^{6} \cdot 3^{3})^{1/3} = 4^{2} \cdot 3 = 16 \). 2. **For the second equation:** \[ 4^{2} \cdot x^{2} \cdot 3^{2} = 84^{2} \] First, compute \( 84^{2} = 7056 \). We substitute \( x = 16 \): \[ 4^{2} \cdot 16^{2} \cdot 3^{2} = 7056 \] Hence: \[ 16 = 4^{2} \cdot 16^{2} \cdot 3^{2} = 4^{2} \cdot 256 \cdot 9 = 1024 \cdot 9 = 9216 \text{ (not equaling 7056)} \] Therefore, let's use \( x \) directly since we want to solve for \( x \). Further simplifying gives LHS \( 16x^{2} = 7056 \), thus \( x^{2} = 441 \), leading to \( x = 21 \). 3. **For the third equation:** \[ 3^{2} \cdot 3^{5} \cdot 3^{x} = 3^{11} \] Summing the exponents gives: \[ 3^{7+x} = 3^{11} \implies 7+x = 11 \implies x = 4. \] 4. **For the fourth equation:** \[ (x^{2})^{5} = 1024 \] Simplifying gives \( x^{10} = 1024 \). Noting that \( 1024 = 2^{10} \) means \( x^{10} = 2^{10} \), hence \( x = 2 \). Putting the derived values together: 1. \( x = 16 \) 2. \( x = 21 \) 3. \( x = 4 \) 4. \( x = 2 \) So there you go! Four equations solved with some number crunching! 🎉
