The number of real roots of the equation
`5+|2^(x)-1|=2^(x)(2^(x)-2)` is

Explanation

To find the number of real roots of the equation

\[ 5 + |2^x - 1| = 2^x(2^x - 2), \]

we will analyze the equation by considering two cases based on the expression inside the absolute value.

### Step 1: Analyze the cases for the absolute value

We will consider two cases based on the expression \(2^x - 1\):

1. **Case 1**: \(2^x - 1 \geq 0\) (i.e., \(x \geq 0\))
2. **Case 2**: \(2^x - 1 < 0\) (i.e., \(x < 0\))

### Step 2: Solve Case 1

In Case 1, since \(2^x - 1 \geq 0\), we can replace \(|2^x - 1|\) with \(2^x - 1\). The equation becomes:

\[ 5 + (2^x - 1) = 2^x(2^x - 2). \]

Simplifying this gives:

\[ 5 + 2^x - 1 = 2^x(2^x - 2), \]
\[ 4 + 2^x = 2^{2x} - 2^{x + 1}. \]

Rearranging leads to:

\[ 2^{2x} - 2^{x + 1} - 2^x - 4 = 0. \]

Let \(t = 2^x\). Then we can rewrite the equation as:

\[ t^2 - 2t - 4 = 0. \]

### Step 3: Solve the quadratic equation

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

\[ t = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}, \]
\[ t = \frac{2 \pm \sqrt{4 + 16}}{2}, \]
\[ t = \frac{2 \pm \sqrt{20}}{2}, \]
\[ t = \frac{2 \pm 2\sqrt{5}}{2}, \]
\[ t = 1 \pm \sqrt{5}. \]

Since \(t = 2^x\) must be positive, we only consider \(t = 1 + \sqrt{5}\).

### Step 4: Find \(x\)

Now we find \(x\):

\[ 2^x = 1 + \sqrt{5} \implies x = \log_2(1 + \sqrt{5}). \]

### Step 5: Solve Case 2

In Case 2, since \(2^x - 1 < 0\), we replace \(|2^x - 1|\) with \(-(2^x - 1)\):

\[ 5 - (2^x - 1) = 2^x(2^x - 2). \]

This simplifies to:

\[ 5 - 2^x + 1 = 2^{2x} - 2^{x + 1}, \]
\[ 6 - 2^x = 2^{2x} - 2^{x + 1}. \]

Rearranging gives:

\[ 2^{2x} - 2^{x + 1} + 2^x - 6 = 0. \]

Let \(m = 2^x\):

\[ m^2 - 2m + m - 6 = 0 \implies m^2 - m - 6 = 0. \]

### Step 6: Solve the quadratic equation for Case 2

Using the quadratic formula again:

\[ m = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm 5}{2}. \]

This gives:

\[ m = 3 \quad \text{or} \quad m = -2. \]

Since \(m = 2^x\) must be positive, we take \(m = 3\).

### Step 7: Find \(x\) for Case 2

\[ 2^x = 3 \implies x = \log_2(3). \]

### Step 8: Combine results

From both cases, we have two solutions:

1. \(x = \log_2(1 + \sqrt{5})\) from Case 1.
2. \(x = \log_2(3)\) from Case 2.

### Conclusion

Thus, the number of real roots of the equation is:

\[ \boxed{2}. \]