How many values of x exist if `4^(x)-12.2^(x)-64=0` ?

Options

A. 8
B. 4
C. 2
D. 1

Answer

To solve the equation \(4^x - 12 \cdot 2^x - 64 = 0\), we will follow these steps:

### Step 1: Rewrite the equation
We start by rewriting \(4^x\) in terms of \(2^x\):
\[ 4^x = (2^2)^x = 2^{2x} \]
Thus, the equation becomes:
\[ 2^{2x} - 12 \cdot 2^x - 64 = 0 \]

### Step 2: Substitute \(y = 2^x\)
Let \(y = 2^x\). Then, \(2^{2x} = y^2\). The equation now transforms into:
\[ y^2 - 12y - 64 = 0 \]

### Step 3: Solve the quadratic equation
Now we will solve the quadratic equation \(y^2 - 12y - 64 = 0\) using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 1\), \(b = -12\), and \(c = -64\).

Calculating the discriminant:
\[ b^2 - 4ac = (-12)^2 - 4 \cdot 1 \cdot (-64) = 144 + 256 = 400 \]
Now substituting back into the quadratic formula:
\[ y = \frac{12 \pm \sqrt{400}}{2 \cdot 1} = \frac{12 \pm 20}{2} \]

### Step 4: Calculate the values of \(y\)
Calculating the two possible values for \(y\):
1. \(y = \frac{12 + 20}{2} = \frac{32}{2} = 16\)
2. \(y = \frac{12 - 20}{2} = \frac{-8}{2} = -4\)

### Step 5: Determine valid values of \(y\)
Since \(y = 2^x\) must be positive, we discard \(y = -4\) because \(2^x\) cannot be negative. Thus, we only consider:
\[ y = 16 \]

### Step 6: Solve for \(x\)
Now we find \(x\) from \(y = 16\):
\[ 2^x = 16 \]
This can be rewritten as:
\[ 2^x = 2^4 \]
Thus, we have:
\[ x = 4 \]

### Conclusion
There is only **one value** of \(x\) that satisfies the original equation.

### Final Answer
The number of values of \(x\) that exist is **1**.

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