If x, y, z are positive real numbers such that `x^(3)y^(2)z^(4)=7`, then the least value of `2x+5y+3z`, is

Options

A. `((525)/(128))^(1//9)`
B. `3((525)/(128))^(1//9)`
C. `9((525)/(128))^(1//9)`
D. none of these

Explanation

To find the least value of \(2x + 5y + 3z\) given the constraint \(x^3 y^2 z^4 = 7\), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-step Solution:

1. **Set up the AM-GM inequality**:
We need to express \(2x + 5y + 3z\) in a way that allows us to apply AM-GM. We can break down the terms based on their coefficients:
\[ 2x = \frac{2x}{3} + \frac{2x}{3} + \frac{2x}{3} \]
\[ 5y = \frac{5y}{2} + \frac{5y}{2} \]
\[ 3z = \frac{3z}{4} + \frac{3z}{4} + \frac{3z}{4} + \frac{3z}{4} \]
This gives us a total of 9 terms:
\[ \frac{2x}{3}, \frac{2x}{3}, \frac{2x}{3}, \frac{5y}{2}, \frac{5y}{2}, \frac{3z}{4}, \frac{3z}{4}, \frac{3z}{4}, \frac{3z}{4} \]

2. **Apply AM-GM**:
By AM-GM, we have:
\[ \frac{\frac{2x}{3} + \frac{2x}{3} + \frac{2x}{3} + \frac{5y}{2} + \frac{5y}{2} + \frac{3z}{4} + \frac{3z}{4} + \frac{3z}{4} + \frac{3z}{4}}{9} \geq \sqrt[9]{\left(\frac{2x}{3}\right)^3 \left(\frac{5y}{2}\right)^2 \left(\frac{3z}{4}\right)^4} \]

3. **Simplify the left-hand side**:
The left-hand side simplifies to:
\[ \frac{2x + 5y + 3z}{9} \]

4. **Calculate the geometric mean**:
The right-hand side becomes:
\[ \sqrt[9]{\left(\frac{2x}{3}\right)^3 \left(\frac{5y}{2}\right)^2 \left(\frac{3z}{4}\right)^4} = \sqrt[9]{\frac{2^3 x^3 \cdot 5^2 y^2 \cdot 3^4 z^4}{3^3 \cdot 2^2 \cdot 4^4}} \]

5. **Substitute the constraint**:
Since \(x^3 y^2 z^4 = 7\), we substitute this into our expression:
\[ \sqrt[9]{\frac{2^3 \cdot 5^2 \cdot 3^4 \cdot 7}{3^3 \cdot 2^2 \cdot 4^4}} \]

6. **Calculate the constants**:
Simplifying the constants:
\[ 4^4 = (2^2)^4 = 2^8 \]
Thus, we have:
\[ \frac{2^3 \cdot 5^2 \cdot 3^4 \cdot 7}{3^3 \cdot 2^2 \cdot 2^8} = \frac{2^{3-2-8} \cdot 5^2 \cdot 3^4 \cdot 7}{3^3} = \frac{2^{-7} \cdot 25 \cdot 81 \cdot 7}{27} \]

7. **Final calculation**:
Now, we can find the minimum value of \(2x + 5y + 3z\):
\[ 2x + 5y + 3z \geq 9 \cdot \sqrt[9]{\frac{2^{-7} \cdot 25 \cdot 81 \cdot 7}{27}} \]

8. **Calculate the minimum value**:
After calculating the above expression, we find that the minimum value of \(2x + 5y + 3z\) is:
\[ \text{Minimum value} = 9 \cdot \sqrt[9]{\frac{525}{128}} = 9 \cdot \frac{525^{1/9}}{2^{7/9}} \]

### Final Result:
The least value of \(2x + 5y + 3z\) is approximately \( \text{value} \).