Solve : `3log_x(4)+ 2log_(4x)4+3log_(16x)4=0`

Explanation

To solve the equation \(3\log_x(4) + 2\log_{4x}(4) + 3\log_{16x}(4) = 0\), we will follow these steps:

### Step 1: Rewrite the logarithmic expressions
Using the property of logarithms that states \(\log_a(b) = \frac{1}{\log_b(a)}\), we can rewrite the logarithmic terms:

\[ 3\log_x(4) = \frac{3}{\log_4(x)}, \quad 2\log_{4x}(4) = \frac{2}{\log_4(4x)}, \quad 3\log_{16x}(4) = \frac{3}{\log_4(16x)} \]

### Step 2: Simplify the logarithmic expressions
Now, we can simplify \(\log_4(4x)\) and \(\log_4(16x)\):

\[ \log_4(4x) = \log_4(4) + \log_4(x) = 1 + \log_4(x) \]
\[ \log_4(16x) = \log_4(16) + \log_4(x) = 2 + \log_4(x) \quad \text{(since } 16 = 4^2\text{)} \]

### Step 3: Substitute back into the equation
Substituting these back into the equation gives us:

\[ \frac{3}{\log_4(x)} + \frac{2}{1 + \log_4(x)} + \frac{3}{2 + \log_4(x)} = 0 \]

### Step 4: Let \(t = \log_4(x)\)
Let \(t = \log_4(x)\). Then, we can rewrite the equation as:

\[ \frac{3}{t} + \frac{2}{1 + t} + \frac{3}{2 + t} = 0 \]

### Step 5: Find a common denominator
The common denominator for the fractions is \(t(1 + t)(2 + t)\). Thus, we can rewrite the equation:

\[ 3(1 + t)(2 + t) + 2t(2 + t) + 3t(1 + t) = 0 \]

### Step 6: Expand the equation
Expanding each term:

1. \(3(1 + t)(2 + t) = 3(2 + 3t + t^2) = 6 + 9t + 3t^2\)
2. \(2t(2 + t) = 4t + 2t^2\)
3. \(3t(1 + t) = 3t + 3t^2\)

Combining these gives:

\[ (3t^2 + 2t^2 + 3t^2) + (9t + 4t + 3t) + 6 = 0 \]

This simplifies to:

\[ 8t^2 + 16t + 6 = 0 \]

### Step 7: Divide the equation by 2
To simplify, divide the entire equation by 2:

\[ 4t^2 + 8t + 3 = 0 \]

### Step 8: Factor the quadratic equation
Now we need to factor or use the quadratic formula to solve for \(t\):

Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = 4\), \(b = 8\), and \(c = 3\):

\[ t = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 4 \cdot 3}}{2 \cdot 4} = \frac{-8 \pm \sqrt{64 - 48}}{8} = \frac{-8 \pm \sqrt{16}}{8} = \frac{-8 \pm 4}{8} \]

Calculating the two possible values for \(t\):

1. \(t = \frac{-4}{8} = -\frac{1}{2}\)
2. \(t = \frac{-12}{8} = -\frac{3}{2}\)

### Step 9: Solve for \(x\)
Recall that \(t = \log_4(x)\):

1. For \(t = -\frac{1}{2}\):
\[ \log_4(x) = -\frac{1}{2} \implies x = 4^{-\frac{1}{2}} = \frac{1}{\sqrt{4}} = \frac{1}{2} \]

2. For \(t = -\frac{3}{2}\):
\[ \log_4(x) = -\frac{3}{2} \implies x = 4^{-\frac{3}{2}} = \frac{1}{4^{3/2}} = \frac{1}{(2^2)^{3/2}} = \frac{1}{2^3} = \frac{1}{8} \]

### Final Solution
Thus, the solutions for \(x\) are:

\[ x = \frac{1}{2} \quad \text{and} \quad x = \frac{1}{8} \]