Understanding the Problem: Exponents, GP, and AP

We are given an equation relating three variables \(p\), \(q\), and \(r\) through exponents: \(p^x = q^y = r^z\). We are also told that the exponents \(x\), \(y\), and \(z\) are in a Geometric Progression (GP). Our task is to determine which of the two given statements about \(p\), \(q\), \(r\) and their logarithms is correct.

Analyzing the Given Information

Let's represent the common relationship \(p^x = q^y = r^z\) by a constant \(k\). So, we have:

• \(p^x = k \implies p = k^{1/x}\)
• \(q^y = k \implies q = k^{1/y}\)
• \(r^z = k \implies r = k^{1/z}\)

We are also given that \(x\), \(y\), and \(z\) are in Geometric Progression (GP). This means the ratio between consecutive terms is constant, i.e., \(\frac{y}{x} = \frac{z}{y}\). Cross-multiplying gives us the condition for \(x, y, z\) being in GP: \(y^2 = xz\).

Evaluating Statement I: \(p\), \(q\), \(r\) are in AP

For \(p\), \(q\), and \(r\) to be in Arithmetic Progression (AP), the difference between consecutive terms must be constant. This means \(q - p = r - q\), or \(2q = p + r\). Let's substitute the expressions for \(p\), \(q\), and \(r\) in terms of \(k\):

We have \(p = k^{1/x}\), \(q = k^{1/y}\), \(r = k^{1/z}\). From \(y^2 = xz\), we can take the reciprocal of both sides: \(\frac{1}{y^2} = \frac{1}{xz}\). This can be rewritten as \(\left(\frac{1}{y}\right)^2 = \frac{1}{x} \cdot \frac{1}{z}\). This implies that the terms \(\frac{1}{x}\), \(\frac{1}{y}\), and \(\frac{1}{z}\) are in Geometric Progression (GP).

Now let's consider the terms \(p\), \(q\), \(r\). Using the property \(q = k^{1/y}\), we can write:

\(q^2 = (k^{1/y})^2 = k^{2/y}\)

Since \(\frac{1}{y}\) is the geometric mean of \(\frac{1}{x}\) and \(\frac{1}{z}\), we have \(\frac{2}{y} = \frac{1}{x} + \frac{1}{z}\) is NOT necessarily true. However, we found that \((\frac{1}{y})^2 = \frac{1}{x} \cdot \frac{1}{z}\). So, \(q^2 = k^{2/y}\). Let's relate this to \(p\) and \(r\). \(p \cdot r = k^{1/x} \cdot k^{1/z} = k^{(1/x + 1/z)}\). We know \(y^2 = xz\). Let's rewrite the exponents: \(p = k^{1/x}\) \(q = k^{1/y}\) \(r = k^{1/z}\) Since \(1/x, 1/y, 1/z\) are in GP, let \(1/y = \lambda (1/x)\) and \(1/z = \lambda (1/y) = \lambda^2 (1/x)\) for some common ratio \(\lambda\). Then \(p = k^{1/x}\) \(q = k^{\lambda/x} = (k^{1/x})^\lambda = p^\lambda\) \(r = k^{\lambda^2/x} = (k^{1/x})^{\lambda^2} = p^{\lambda^2}\) This shows that \(p, q, r\) themselves form a GP (\(q/p = r/q\)), not an AP (\(q-p = r-q\)). Therefore, Statement I is incorrect.

Evaluating Statement II: \(\ln p\), \(\ln q\), \(\ln r\) are in GP

Let's find the natural logarithms (\(\ln\)) of \(p\), \(q\), and \(r\). We assume \(k > 0\) and \(k \neq 1\) for the logarithms to be well-defined and non-zero.

• \(\ln p = \ln(k^{1/x}) = \frac{1}{x} \ln k\)
• \(\ln q = \ln(k^{1/y}) = \frac{1}{y} \ln k\)
• \(\ln r = \ln(k^{1/z}) = \frac{1}{z} \ln k\)

For these three logarithmic terms to be in Geometric Progression (GP), the square of the middle term must equal the product of the other two terms:

\((\ln q)^2 = (\ln p)(\ln r)\)

Substitute the expressions:

\(\left(\frac{1}{y} \ln k\right)^2 = \left(\frac{1}{x} \ln k\right) \left(\frac{1}{z} \ln k\right)\)

\(\frac{1}{y^2} (\ln k)^2 = \frac{1}{xz} (\ln k)^2\)

Assuming \(\ln k \neq 0\) (i.e., \(k \neq 1\)), we can divide both sides by \((\ln k)^2\): \(\frac{1}{y^2} = \frac{1}{xz}\)

This simplifies to \(y^2 = xz\). We know from the problem statement that \(x\), \(y\), and \(z\) are in GP, which means \(y^2 = xz\) is true. Therefore, Statement II is correct.

Conclusion

Based on the analysis, Statement I is incorrect because \(p\), \(q\), \(r\) are in GP, not AP. Statement II is correct because the condition \(y^2 = xz\) directly leads to \(\ln p\), \(\ln q\), \(\ln r\) being in GP.

Thus, only Statement II is correct.