If `9^(x)-2.3^(x)-3=0` , what is the value of x?

Answer

To solve the equation \( 9^x - 2 \cdot 3^x - 3 = 0 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 9^x - 2 \cdot 3^x - 3 = 0 \]
We know that \( 9 \) can be expressed as \( 3^2 \). Therefore, we can rewrite \( 9^x \) as \( (3^2)^x = 3^{2x} \). Substituting this into the equation gives us:
\[ 3^{2x} - 2 \cdot 3^x - 3 = 0 \]

### Step 2: Substitute \( y = 3^x \)
Let \( y = 3^x \). Then \( 3^{2x} = y^2 \). Substituting \( y \) into the equation, we have:
\[ y^2 - 2y - 3 = 0 \]

### Step 3: Factor the quadratic equation
Next, we need to factor the quadratic equation:
\[ y^2 - 2y - 3 = 0 \]
This can be factored as:
\[ (y - 3)(y + 1) = 0 \]

### Step 4: Solve for \( y \)
Setting each factor to zero gives us:
1. \( y - 3 = 0 \) → \( y = 3 \)
2. \( y + 1 = 0 \) → \( y = -1 \)

### Step 5: Substitute back for \( y \)
Recall that \( y = 3^x \). Therefore, we substitute back:
1. From \( y = 3 \):
\[ 3^x = 3 \]
This implies \( x = 1 \) (since \( 3^1 = 3 \)).

2. From \( y = -1 \):
\[ 3^x = -1 \]
This is not possible since \( 3^x \) is always positive for real \( x \).

### Conclusion
The only valid solution is:
\[ x = 1 \]

Thus, the value of \( x \) is \( \boxed{1} \).

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