Solve log₄(9x) + log₄(x+4) - log₄(3) = ln(2e) + ln(1/2e): Finding X

$\log_49x+\log_4(x+4)-\log_43=\ln2e+\ln\frac{1}{2e}$

Find X

To solve this logarithmic equation, we will simplify both sides using logarithm properties.

Step 1: Combine the logarithms on the left side.

The left side is $\log_4 9x + \log_4 (x+4) - \log_4 3$. Using the properties of logarithms, we can combine these logs:

$\log_4 \left( \frac{9x(x+4)}{3} \right)$

This simplifies to:

$\log_4 \left(3x(x+4)\right)$

Step 2: Simplify the right side.

The right side is $\ln 2e + \ln \frac{1}{2e}$. Using properties of natural logarithms, combine as follows:

$\ln \left(2e \cdot \frac{1}{2e}\right) = \ln 1 = 0$

Step 3: Equating both sides, we have:

$\log_4 \left(3x(x+4)\right) = 0$

Step 4: Convert the logarithmic equation to an exponential equation. Since the logarithmic expression equals zero, it signifies:

$3x(x+4) = 4^0 = 1$

Step 5: Solve the equation $3x(x+4) = 1$:

Combine and expand the terms:

$3x^2 + 12x - 1 = 0$

Step 6: Solve the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 3$, $b = 12$, and $c = -1$:

$x = \frac{-12 \pm \sqrt{12^2 - 4 \times 3 \times (-1)}}{2 \times 3}$

Calculate:

$x = \frac{-12 \pm \sqrt{144 + 12}}{6}$

$x = \frac{-12 \pm \sqrt{156}}{6}$

$x = \frac{-12 \pm \sqrt{4 \times 39}}{6}$

$x = \frac{-12 \pm 2\sqrt{39}}{6}$

$x = \frac{-6 \pm \sqrt{39}}{3}$

Thus, the solution is:

$x = -2 + \frac{\sqrt{39}}{3}$

This matches the correct choice.

Therefore, the solution to the problem is $-2+\frac{\sqrt{39}}{3}$.