If `"log"_(a) ab = x,` then the value of `"log"_(b)ab,` is

Options

A. `(x-1)/(x)`
B. `(x)/(x-1)`
C. `(x)/(x+1)`
D. `(x+1)/(x)`

Answer

To solve the problem, we start with the given equation:

**Given:**
\[ \log_a ab = x \]

We need to find the value of:
\[ \log_b ab \]

**Step 1: Rewrite the logarithm using properties**
Using the property of logarithms, we can express \(\log_a ab\) as:
\[ \log_a ab = \log_a a + \log_a b \]
Since \(\log_a a = 1\), we have:
\[ \log_a ab = 1 + \log_a b \]
Thus, we can equate this to \(x\):
\[ 1 + \log_a b = x \]

**Step 2: Solve for \(\log_a b\)**
Rearranging the above equation gives us:
\[ \log_a b = x - 1 \]

**Step 3: Change of base formula**
We want to find \(\log_b ab\). Using the change of base formula, we can express \(\log_b a\) in terms of \(\log_a b\):
\[ \log_b a = \frac{1}{\log_a b} \]
Substituting the value of \(\log_a b\) from Step 2:
\[ \log_b a = \frac{1}{x - 1} \]

**Step 4: Find \(\log_b ab\)**
Now, we can express \(\log_b ab\) as:
\[ \log_b ab = \log_b a + \log_b b \]
Since \(\log_b b = 1\), we have:
\[ \log_b ab = \log_b a + 1 = \frac{1}{x - 1} + 1 \]

**Step 5: Simplify the expression**
To combine the terms, we need a common denominator:
\[ \log_b ab = \frac{1}{x - 1} + \frac{x - 1}{x - 1} = \frac{1 + (x - 1)}{x - 1} = \frac{x}{x - 1} \]

Thus, the final answer is:
\[ \log_b ab = \frac{x}{x - 1} \]

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