Answer:
Step-by-step explanation:
Given:
\[ \log_h(ab) = m \]
\[ \log_a(ab) = \frac{m}{m - 1} \]
We know that \(\log_{b} (x) = \frac{1}{\log_{x} (b)}\). We can use this property to rewrite the first equation:
\[ \log_h(ab) = \frac{1}{\log_{ab} (h)} = m \]
Since \(\log_{ab} (h)\) is equal to \(\frac{1}{\log_{h} (ab)}\), we can rewrite the first equation as:
\[ \log_h(ab) = \frac{1}{\frac{1}{\log_{h} (ab)}} = m \]
This simplifies to:
\[ \log_{h} (ab) = \frac{1}{m} \]
Now, let's use the second equation:
\[ \log_a(ab) = \frac{m}{m - 1} \]
We can rewrite this using the property \(\log_{b} (x) = \frac{1}{\log_{x} (b)}\):
\[ \log_{ab} (a^b) = \frac{m}{m - 1} \]
\[ \frac{b}{\log_{a} (ab)} = \frac{m}{m - 1} \]
\[ \frac{\log_{a} (ab)}{b} = \frac{m - 1}{m} \]
\[ \log_{a} (ab) = \frac{b(m - 1)}{m} \]
Now, we know that:
\[ \log_{h} (ab) = \frac{1}{m} \]
\[ \log_{a} (ab) = \frac{b(m - 1)}{m} \]
Using the fact that \(\log_{h} (ab) = \frac{1}{\log_{ab} (h)}\), we can substitute \(\frac{1}{m}\) for \(\log_{h} (ab)\) in the equation for \(\log_{a} (ab)\):
\[ \frac{1}{\frac{1}{m}} = \frac{b(m - 1)}{m} \]
\[ m = \frac{b(m - 1)}{m} \]
\[ m^2 = b(m - 1) \]
\[ m^2 = bm - b \]
\[ m^2 - bm + b = 0 \]
This is a quadratic equation in terms of \(m\). We can solve it using the quadratic formula:
\[ m = \frac{-(-b) \pm \sqrt{(-b)^2 - 4 \cdot 1 \cdot b}}{2 \cdot 1} \]
\[ m = \frac{b \pm \sqrt{b^2 - 4b}}{2} \]
Since \(m\) cannot be negative, we take the positive root:
\[ m = \frac{b + \sqrt{b^2 - 4b}}{2} \]
So, the value of \(m\) in terms of \(b\) is \(m = \frac{b + \sqrt{b^2 - 4b}}{2}\).
