To make the bases common in the equation \( 6y^3 = (\sqrt[3]{4})^x \), we need to express both sides with a common base.
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Rewrite the left side: \(6y^3\) can be expressed in terms of its prime factors: \[ 6 = 2 \cdot 3 \] So, \[ 6y^3 = 2 \cdot 3 \cdot y^3 \]
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Rewrite the right side: The expression \((\sqrt[3]{4})^x\) can be rewritten using exponents: \[ \sqrt[3]{4} = 4^{1/3} = (2^2)^{1/3} = 2^{2/3} \] Thus, \[ (\sqrt[3]{4})^x = (2^{2/3})^x = 2^{(2/3)x} \]
Now we have: \[ 2 \cdot 3 \cdot y^3 = 2^{(2/3)x} \]
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Distribute common bases: To balance the equation, we need to match the bases:
- On the left side, we have \(2\) from \(6\).
- Therefore, we must separate the \(3\) and \(y^3\) into their respective bases.
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Rearrange the equation according to powers of base \(2\):
From the left side: \[ 2^1 \cdot (3y^3) = 2^{(2/3)x} \] implies that we need to equalize the powers of base \(2\).
Clearly, the only common base extracted is \(2\) on the left-hand side, while \(3y^3\) remains uncombined.
- Since there is no common base for the complete left-hand side with the right-hand side, it indicates balancing directly isn't viable only on the basis of power manipulation.
The equation can be expressed, but preserving the equality would need different treatments like logarithms or a numerical approach depending on \(y\).
For explicit methods such as: \[ 3y^3 = \frac{2^{(2/3)x}}{2^1} \]
In logarithmic form, considering balances on the powers become possible but lead to equation transfiguration.
To summarize, while we were unable to completely unify or balance in straightforward terms under a prime base, we can still compare or analyze between \(2\) and \(y^3\) respectively utilizing numerical methods, logarithmic expressions, or further case study beyond mere factored representation.
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