We are given that:

$\log_3 y = x \quad \text{and} \quad \log_2 z = x$

We need to find what $72^x$ is equal to in terms of $y$ and $z$.

Step 1: Expressing $y$ and $z$ in terms of $x$

From $\log_3 y = x$, we can express $y$ as: $y = 3^x$

From $\log_2 z = x$, we can express $z$ as: $z = 2^x$

Step 2: Breaking down $72^x$

We know that: $72 = 2^3 \times 3^2$

So, $72^x$ can be written as: $72^x = (2^3 \times 3^2)^x = 2^{3x} \times 3^{2x}$

Step 3: Substituting $y = 3^x$ and $z = 2^x$

We can now substitute $y = 3^x$ and $z = 2^x$ into the expression: $72^x = 2^{3x} \times 3^{2x}$

Notice that $2^{3x} = (2^x)^3 = z^3$ and $3^{2x} = (3^x)^2 = y^2$. Therefore, we have: $72^x = z^3 \times y^2$

Thus, $72^x = y^2 z^3$.

Answer:

The correct option is (c): $y^2 z^3$.

Would you like more details or have any other questions? Here are 5 related questions to expand on this topic:

1. How would the solution change if the logarithmic bases were different?
2. Can you prove why $\log_b a = x$ implies $a = b^x$?
3. What is the general form of simplifying powers of products like $(a \cdot b)^n$?
4. How would the result change if the exponent was $x^2$ instead of $x$?
5. How does the relationship between logarithms and exponents help in simplifying expressions?

Tip: Remember that logarithmic identities can often be used to convert between multiplication and exponentiation, making it easier to simplify complex expressions!