Question

make the bases comon 6y^(3)=( root(3)(4))^(x)

Answer 1

To make the bases common in the equation \( 6y^3 = (\sqrt[3]{4})^x \), we need to express both sides with a common base.

1. **Rewrite the left side**: \(6y^3\) can be expressed in terms of its prime factors:
\[
6 = 2 \cdot 3
\]
So,
\[
6y^3 = 2 \cdot 3 \cdot y^3
\]

2. **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:
\[
2 \cdot 3 \cdot y^3 = 2^{(2/3)x}
\]

3. **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.

4. **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.

5. **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.

Let me know if you need further assistance!