If `f (x) = log_(x^2) 36 and g(x) = log_x 6`, then f (x) = g(x) holds for x belonging to

Options

A. R
B. ` phi`
C. `(0,oo)-{1}`
D. None

Answer

To solve the equation \( f(x) = g(x) \) where \( f(x) = \log_{x^2} 36 \) and \( g(x) = \log_x 6 \), we will follow these steps:

### Step 1: Rewrite the logarithmic expressions
We can use the change of base formula for logarithms, which states that \( \log_a b = \frac{\log_c b}{\log_c a} \) for any positive \( c \).

So, we can rewrite \( f(x) \) and \( g(x) \):
\[ f(x) = \log_{x^2} 36 = \frac{\log 36}{\log x^2} = \frac{\log 36}{2 \log x} \]
\[ g(x) = \log_x 6 = \frac{\log 6}{\log x} \]

### Step 2: Set the two expressions equal to each other
Now, we set \( f(x) \) equal to \( g(x) \):
\[ \frac{\log 36}{2 \log x} = \frac{\log 6}{\log x} \]

### Step 3: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ \log 36 \cdot \log x = 2 \log 6 \]

### Step 4: Simplify the equation
We know that \( \log 36 = \log(6^2) = 2 \log 6 \). Substituting this into the equation gives:
\[ 2 \log 6 \cdot \log x = 2 \log 6 \]

### Step 5: Divide both sides by \( 2 \log 6 \)
Assuming \( \log 6 \neq 0 \) (which is true since \( 6 > 1 \)), we can divide both sides by \( 2 \log 6 \):
\[ \log x = 1 \]

### Step 6: Solve for \( x \)
Now, we can solve for \( x \):
\[ x = 10^1 = 10 \]

### Step 7: Check the conditions for the logarithm
We need to ensure that the base of the logarithm is valid. The base \( x \) must be greater than 0 and not equal to 1. Since \( x = 10 \), it satisfies these conditions.

### Conclusion
Thus, the solution to the equation \( f(x) = g(x) \) holds for \( x = 10 \).