If `3^(2x+1)*4^(x-1)=36` then find the value of x

Options

A. `"log"_(36)48`
B. `"log"_(48)36`
C. `"log"_(24)12`
D. `"log"_(12)24`

Answer

To solve the equation \( 3^{2x+1} \cdot 4^{x-1} = 36 \), we can follow these steps:

### Step 1: Rewrite the equation
We start by rewriting \( 3^{2x+1} \) and \( 4^{x-1} \):
\[ 3^{2x+1} = 3^{2x} \cdot 3^1 = 3^{2x} \cdot 3 \]
\[ 4^{x-1} = \frac{4^x}{4} = \frac{4^x}{2^2} = \frac{(2^2)^x}{2^2} = \frac{2^{2x}}{4} \]
Thus, we can rewrite the original equation as:
\[ 3^{2x} \cdot 3 \cdot \frac{2^{2x}}{4} = 36 \]

### Step 2: Simplify the equation
Now, we simplify the left-hand side:
\[ \frac{3^{2x} \cdot 2^{2x} \cdot 3}{4} = 36 \]
Multiply both sides by 4:
\[ 3^{2x} \cdot 2^{2x} \cdot 3 = 144 \]

### Step 3: Combine the terms
We can combine \( 3^{2x} \cdot 2^{2x} \) as follows:
\[ 3^{2x} \cdot 2^{2x} = (3 \cdot 2)^{2x} = 6^{2x} \]
Thus, we rewrite the equation:
\[ 6^{2x} \cdot 3 = 144 \]

### Step 4: Isolate \( 6^{2x} \)
Now, divide both sides by 3:
\[ 6^{2x} = \frac{144}{3} = 48 \]

### Step 5: Take logarithm of both sides
Taking logarithm on both sides:
\[ \log(6^{2x}) = \log(48) \]
Using the property of logarithms \( \log(a^b) = b \log(a) \):
\[ 2x \log(6) = \log(48) \]

### Step 6: Solve for \( x \)
Now, isolate \( x \):
\[ x = \frac{\log(48)}{2 \log(6)} \]

### Step 7: Convert to a different logarithmic base
Using the change of base formula, we can express this as:
\[ x = \frac{1}{2} \cdot \frac{\log(48)}{\log(6)} = \frac{1}{2} \log_6(48) \]

### Final Answer
Thus, the value of \( x \) is:
\[ x = \frac{1}{2} \log_6(48) \]

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