If $$ \log 3 $$, $$ \log (3^x - 2) $$, and $$ \log (3^x + 4) $$ are in arithmetic progression, find the value of $$ x $$.

Question

Accepted Answer

Given that $$ \log 3 $$, $$ \log (3^x - 2) $$, and $$ \log (3^x + 4) $$ are in arithmetic progression, the middle term is the average of the other two terms. Thus, we can write:  
$$
\log (3^x - 2) = \frac{\log 3 + \log (3^x + 4)}{2}.
$$

Simplify the right-hand side:
$$
\log (3^x - 2) = \frac{\log [3(3^x + 4)]}{2}.
$$

Using logarithmic properties:
$$
\log (3^x - 2) = \frac{\log [3^{x+1} + 12]}{2}.
$$

Equating both sides:
$$
3^x - 2 = \sqrt{3^{x+1} + 12}.
$$

Squaring both sides:
$$
(3^x - 2)^2 = 3^{x+1} + 12.
$$

Expand and simplify:
$$
9^x - 4(3^x) + 4 = 3^{x+1} + 12.
$$

Substitute $$ 9^x = (3^x)^2 $$:
$$
(3^x)^2 - 4(3^x) + 4 = 3^{x+1} + 12.
$$

Replace $$ 3^{x+1} = 3(3^x) $$:
$$
(3^x)^2 - 4(3^x) + 4 = 3(3^x) + 12.
$$

Simplify further:
$$
(3^x)^2 - 7(3^x) - 8 = 0.
$$

Let $$ y = 3^x $$. The equation becomes:
$$
y^2 - 7y - 8 = 0.
$$

Solve this quadratic equation using the quadratic formula:
$$
y = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(-8)}}{2(1)}.
$$
$$
y = \frac{7 \pm \sqrt{49 + 32}}{2}.
$$
$$
y = \frac{7 \pm \sqrt{81}}{2}.
$$
$$
y = \frac{7 + 9}{2} \quad 	ext{or} \quad y = \frac{7 - 9}{2}.
$$
$$
y = 8 \quad 	ext{or} \quad y = -1.
$$

Since $$ y = 3^x > 0 $$, we have $$ y = 8 $$.  
Thus, $$ 3^x = 8 $$, and taking logarithm base $$ 3 $$:
$$
x = \log_3(8).
$$

Answer

Answer

Answer