Find the number of real values of x satisfying the equation.
`log_(2)(4^(x+1)+4)*log_(2)(4^(x)+1)=log_(1//sqrt(2)) sqrt((1)/(8))`

Answer

To solve the equation

\[ \log_{2}(4^{(x+1)} + 4) \cdot \log_{2}(4^{x} + 1) = \log_{\frac{1}{\sqrt{2}}} \sqrt{\frac{1}{8}}, \]

we will follow these steps:

### Step 1: Simplify the Right Side
First, we simplify the right-hand side of the equation.

\[ \log_{\frac{1}{\sqrt{2}}} \sqrt{\frac{1}{8}} = \log_{\frac{1}{\sqrt{2}}} \left( \frac{1}{8} \right)^{\frac{1}{2}} = \log_{\frac{1}{\sqrt{2}}} \frac{1}{\sqrt{8}}. \]

Next, we can express \(\frac{1}{\sqrt{8}}\) as \(\frac{1}{2^{3/2}} = 2^{-3/2}\). Thus,

\[ \log_{\frac{1}{\sqrt{2}}} \left( 2^{-3/2} \right) = -\frac{3}{2} \cdot \log_{\frac{1}{\sqrt{2}}} 2. \]

Since \(\log_{\frac{1}{\sqrt{2}}} 2 = -1\) (because \(\frac{1}{\sqrt{2}} = 2^{-1/2}\)), we have:

\[ \log_{\frac{1}{\sqrt{2}}} \sqrt{\frac{1}{8}} = -\frac{3}{2} \cdot (-1) = \frac{3}{2}. \]

### Step 2: Rewrite the Equation
Now, we rewrite the original equation:

\[ \log_{2}(4^{(x+1)} + 4) \cdot \log_{2}(4^{x} + 1) = \frac{3}{2}. \]

### Step 3: Change of Base
Next, we can express \(4\) as \(2^2\):

\[ \log_{2}(4^{(x+1)} + 4) = \log_{2}(2^{2(x+1)} + 2^2) = \log_{2}(2^{2x + 2} + 4). \]

### Step 4: Set \(t = \log_{2}(4^{x} + 1)\)
Let \(t = \log_{2}(4^{x} + 1)\). Then, we rewrite the equation as:

\[ \log_{2}(2^{2x + 2} + 4) \cdot t = \frac{3}{2}. \]

### Step 5: Expand and Rearrange
Using properties of logarithms, we can express \(t\):

\[ t = \log_{2}(4^{x} + 1) = \log_{2}(2^{2x} + 1). \]

Now we have:

\[ \log_{2}(2^{2x + 2} + 4) \cdot \log_{2}(2^{2x} + 1) = \frac{3}{2}. \]

### Step 6: Solve for \(t\)
We can express the equation in terms of \(t\):

Let \(u = \log_{2}(4^{x} + 1)\):

\[ \log_{2}(4^{(x+1)} + 4) = \log_{2}(4^{x} + 4) = \log_{2}(4(4^{x} + 1)). \]

Thus, we can write:

\[ \log_{2}(4) + \log_{2}(4^{x} + 1) = 2 + t. \]

### Step 7: Substitute Back
Substituting back, we have:

\[ (2 + t)t = \frac{3}{2}. \]

This expands to:

\[ 2t + t^2 = \frac{3}{2}. \]

### Step 8: Rearranging the Quadratic Equation
Rearranging gives:

\[ t^2 + 2t - \frac{3}{2} = 0. \]

### Step 9: Solve the Quadratic Equation
Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\[ t = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot \left(-\frac{3}{2}\right)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 6}}{2} = \frac{-2 \pm \sqrt{10}}{2} = -1 \pm \frac{\sqrt{10}}{2}. \]

### Step 10: Find Real Values of \(x\)
Now we have two possible values for \(t\):

1. \(t_1 = -1 + \frac{\sqrt{10}}{2}\)
2. \(t_2 = -1 - \frac{\sqrt{10}}{2}\)

Since \(t = \log_{2}(4^{x} + 1)\) must be non-negative, we only consider \(t_1\):

\[ -1 + \frac{\sqrt{10}}{2} \geq 0 \implies \sqrt{10} \geq 2 \implies x \text{ is valid.} \]

### Conclusion
Thus, the only valid solution corresponds to \(x = 0\).

The number of real values of \(x\) satisfying the equation is:

\[ \boxed{1}. \]