Logarithm Properties Application

This problem asks us to find the value of the expression \((ace)^{\frac{1}{p}}\) given three logarithmic relationships.

Given Logarithmic Relationships

We are provided with the following equations:

• \(\log_ba = p\)
• \(\log_dc = 2p\)
• \(\log_fe = 3p\)

Our goal is to use these relationships to evaluate \((ace)^{\frac{1}{p}}\).

Exponential Conversion Strategy

The fundamental property of logarithms states that if \(\log_xy = z\), then \(x^z = y\). We will use this property to convert each given logarithmic equation into its equivalent exponential form.

• From \(\log_ba = p\), we can rewrite this in exponential form as: \(a = b^p\)
• From \(\log_dc = 2p\), we get: \(c = d^{2p}\)
• From \(\log_fe = 3p\), we get: \(e = f^{3p}\)

Expression Evaluation Steps

Now, let's substitute these exponential forms back into the expression we need to evaluate: \((ace)^{\frac{1}{p}}\).

1. First, find the product \(ace\): \(ace = (b^p) \cdot (d^{2p}) \cdot (f^{3p})\)
2. Using the exponent rule \(x^m \cdot y^n \cdot z^o = x^m y^n z^o\), the expression remains: \(ace = b^p d^{2p} f^{3p}\)
3. Now, we need to raise this product to the power of \(\frac{1}{p}\): \((ace)^{\frac{1}{p}} = (b^p d^{2p} f^{3p})^{\frac{1}{p}}\)
4. Apply the power of a product rule, which states \((xyz)^k = x^k y^k z^k\), and the power of a power rule \((x^m)^n = x^{mn}\): \((ace)^{\frac{1}{p}} = (b^p)^{\frac{1}{p}} \cdot (d^{2p})^{\frac{1}{p}} \cdot (f^{3p})^{\frac{1}{p}}\)
5. Simplify each term by multiplying the exponents:
   • \((b^p)^{\frac{1}{p}} = b^{p \times \frac{1}{p}} = b^1 = b\)
   • \((d^{2p})^{\frac{1}{p}} = d^{2p \times \frac{1}{p}} = d^2\)
   • \((f^{3p})^{\frac{1}{p}} = f^{3p \times \frac{1}{p}} = f^3\)

Final Result Derivation

Combining the simplified terms, we get the final value of the expression:

Thus, the expression \((ace)^{\frac{1}{p}}\) is equal to \(bd^2f^3\).