Question

question
solve for all values of ( x ).
( log_{2}(2x + 1) + log_{2}(3x + 7) = 1 )

Explanation:

Step1: Combine log terms via product rule

$\log_2[(2x+1)(3x+7)] = 1$

Step2: Convert to exponential form

$(2x+1)(3x+7) = 2^1 = 2$

Step3: Expand left-hand side

$6x^2 + 14x + 3x + 7 = 2$

Step4: Simplify to quadratic equation

$6x^2 + 17x + 5 = 0$

Step5: Factor the quadratic

$(2x+5)(3x+1) = 0$

Step6: Solve for x, check domain

Solutions from factoring: $x = -\frac{5}{2}$, $x = -\frac{1}{3}$. For logarithms, $2x+1>0$ and $3x+7>0$, so $x>-\frac{1}{2}$. Reject $x=-\frac{5}{2}$.

Answer:

$x = -\frac{1}{3}$