Solve the equation for `x : log4+(1+1/(2x))log3=log(3^(1/x)+27)`

Options

A. 0
B. 2
C. 3
D. None of these

Answer

To solve the equation \( \log 4 + \left(1 + \frac{1}{2x}\right) \log 3 = \log \left(3^{\frac{1}{x}} + 27\right) \), we will follow these steps:

### Step 1: Rewrite the Equation
We start with the given equation:
\[ \log 4 + \left(1 + \frac{1}{2x}\right) \log 3 = \log \left(3^{\frac{1}{x}} + 27\right) \]

### Step 2: Apply Logarithmic Properties
Using the properties of logarithms, we can rewrite the left-hand side:
\[ \log 4 + \log 3^{1 + \frac{1}{2x}} = \log \left(4 \cdot 3^{1 + \frac{1}{2x}}\right) \]
This simplifies our equation to:
\[ \log \left(4 \cdot 3^{1 + \frac{1}{2x}}\right) = \log \left(3^{\frac{1}{x}} + 27\right) \]

### Step 3: Remove the Logarithm
Since the logarithms are equal, we can set the arguments equal to each other:
\[ 4 \cdot 3^{1 + \frac{1}{2x}} = 3^{\frac{1}{x}} + 27 \]

### Step 4: Substitute for Simplicity
Let \( t = 3^{\frac{1}{2x}} \). Then \( 3^{\frac{1}{x}} = t^2 \). Substituting this into the equation gives:
\[ 4 \cdot 3 \cdot t = t^2 + 27 \]
which simplifies to:
\[ 12t = t^2 + 27 \]

### Step 5: Rearranging the Equation
Rearranging the equation leads to:
\[ t^2 - 12t + 27 = 0 \]

### Step 6: Factor the Quadratic
We can factor the quadratic:
\[ (t - 9)(t - 3) = 0 \]
Thus, the solutions for \( t \) are:
\[ t = 9 \quad \text{or} \quad t = 3 \]

### Step 7: Back Substitute for \( x \)
Recall that \( t = 3^{\frac{1}{2x}} \):
1. For \( t = 9 \):
\[ 3^{\frac{1}{2x}} = 9 \implies 3^{\frac{1}{2x}} = 3^2 \implies \frac{1}{2x} = 2 \implies 2x = \frac{1}{2} \implies x = \frac{1}{4} \]

2. For \( t = 3 \):
\[ 3^{\frac{1}{2x}} = 3 \implies \frac{1}{2x} = 1 \implies 2x = 1 \implies x = \frac{1}{2} \]

### Step 8: Check Validity of Solutions
We need to check if these values of \( x \) are valid. The logarithm \( \log \left(3^{\frac{1}{x}} + 27\right) \) requires \( 3^{\frac{1}{x}} + 27 > 0 \), which is satisfied for both \( x = \frac{1}{4} \) and \( x = \frac{1}{2} \).

### Conclusion
Thus, the solutions to the equation are:
\[ x = \frac{1}{4} \quad \text{and} \quad x = \frac{1}{2} \]