Question

solve for x
log2x (2x^2+8x-6)=2

Answer 1

To solve the equation \( \log_{2x}(2x^2 + 8x - 6) = 2 \), we follow these steps:

1. **Change the logarithmic equation to its exponential form:**
\[
2x \text{ raised to the power of } 2 = 2x^2 + 8x - 6
\]

This translates to:
\[
(2x)^2 = 2x^2 + 8x - 6
\]

2. **Calculate \( (2x)^2 \):**
\[
4x^2 = 2x^2 + 8x - 6
\]

3. **Rearrange the equation:**
Subtract \( 2x^2 + 8x - 6 \) from both sides:
\[
4x^2 - 2x^2 - 8x + 6 = 0
\]
Simplifying gives:
\[
2x^2 - 8x + 6 = 0
\]

4. **Simplify further by dividing the whole equation by 2:**
\[
x^2 - 4x + 3 = 0
\]

5. **Factor the quadratic equation:**
\[
(x - 3)(x - 1) = 0
\]

6. **Set each factor to zero and solve for \( x \):**
\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\]
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]

7. **Check if both solutions are valid by substituting back into the original conditions:**

**For \( x = 3 \):**
\[
2x = 2(3) = 6 \text{ (valid since logarithm bases must be positive and not equal to 1)}
\]
\[
2x^2 + 8x - 6 = 2(3^2) + 8(3) - 6 = 2(9) + 24 - 6 = 18 + 24 - 6 = 36 \text{ (valid)}
\]
\[
\log_{6}(36) = 2 \text{ (valid since } 6^2 = 36\text{)}
\]

**For \( x = 1 \):**
\[
2x = 2(1) = 2 \text{ (valid)}
\]
\[
2x^2 + 8x - 6 = 2(1^2) + 8(1) - 6 = 2 + 8 - 6 = 4 \text{ (valid)}
\]
\[
\log_{2}(4) = 2 \text{ (valid since } 2^2 = 4\text{)}
\]

Both solutions \( x = 3 \) and \( x = 1 \) are valid.

Thus, the solutions are:
\[
\boxed{1 \text{ and } 3}
\]