If $\log 3, \log (3^x - 2)$ and $\log (3^x + 4)$ are in arithmetic progression, then $x$ is equal to

### Concept & Formula This problem combines Logarithmic properties with the rule of Arithmetic Progressions (AP). If three terms $A, B, C$ are in an arithmetic progression, the middle term is the average of the other two, which gives the fundamental AP formula: $$ 2B = A + C $$ We will also use the logarithmic power and product rules: 1. $n \log x = \log(x^n)$ 2. $\log x + \log y = \log(xy)$ ### Step-by-Step Solution Given the terms are in AP: $$ 2 \log(3^x - 2) = \log 3 + \log(3^x + 4) $$ Step 1: Apply the log rules to consolidate the equation. Bring the $2$ into the exponent on the left, and multiply the arguments on the right. $$ \log((3^x - 2)^2) = \log(3(3^x + 4)) $$ Step 2: Since the logs are equal, their arguments must be equal. Drop the logarithms. $$ (3^x - 2)^2 = 3(3^x + 4) $$ Step 3: Substitute $y = 3^x$ to make it a standard quadratic equation. $$ (y - 2)^2 = 3(y + 4) $$ Step 4: Expand both sides. $$ y^2 - 4y + 4 = 3y + 12 $$ Step 5: Move all terms to one side to solve the quadratic. $$ y^2 - 7y - 8 = 0 $$ Step 6: Factor the equation. $$ (y - 8)(y + 1) = 0 $$ So, $y = 8$ or $y = -1$. Step 7: Substitute $3^x$ back in for $y$. Since an exponential function ($3^x$) is always positive for real numbers, $3^x = -1$ is impossible and must be rejected. $$ 3^x = 8 $$ Step 8: Solve for $x$ by converting to logarithmic form. $$ x = \log_{3} 8 $$ ### Exam Strategy & Shortcut **Option Verification:** Instead of solving the quadratic, you can test the options directly into $3^x$. If you test option (b) where $x = \log_{3} 8$, then $3^x$ simply becomes $8$. The original sequence becomes: $\log 3, \log(8-2), \log(8+4)$, which simplifies to $\log 3, \log 6, \log 12$. Check the common difference: $\log 6 - \log 3 = \log(6/3) = \log 2$ $\log 12 - \log 6 = \log(12/6) = \log 2$ The common differences match, instantly confirming option (b) is correct. ### Common Pitfall The main trap is forgetting to reject the negative root ($y = -1$). If students don't remember that $a^x > 0$ for positive bases, they might get confused or look for an option that accommodates both roots. Always discard negative exponential outputs. ### Final Answer **Therefore, the correct answer is $\log_{3} 8$.**