Key Concepts and Formulas

• Arithmetic Progression (A.P.): Three numbers $a, b, c$ are in A.P. if the difference between consecutive terms is constant, which means $b - a = c - b$, or equivalently, $2b = a + c$.
• Logarithm Properties:
  • $\log_b a = c \iff b^c = a$
  • $\log_b (MN) = \log_b M + \log_b N$
  • $\log_b \left(\frac{M}{N}\right) = \log_b M - \log_b N$
  • $\log_b M^k = k \log_b M$
  • $\log_{b^k} M = \frac{1}{k} \log_b M$
  • $\log_b b = 1$
  • $b^{\log_b M} = M$
• Domain of Logarithms: For $\log_B A$ to be defined, $B > 0$, $B \ne 1$, and $A > 0$.

Step-by-Step Solution

The given terms are $1$, ${\log _9}\,\,({3^{1 - x}} + 2)$, and ${\log _3}\,\,({4.3^x} - 1)$. Let $a = 1$, $b = {\log _9}\,\,({3^{1 - x}} + 2)$, and $c = {\log _3}\,\,({4.3^x} - 1)$.

Step 1: Apply the Arithmetic Progression Condition Since the terms are in A.P., the middle term is the arithmetic mean of the other two: $2b = a + c$. $2 \cdot {\log _9}\,\,({3^{1 - x}} + 2) = 1 + {\log _3}\,\,({4.3^x} - 1)$ Explanation: This is the fundamental definition of an arithmetic progression applied to the given three terms, setting up the core equation to solve.

Step 2: Simplify the Logarithmic Terms to a Common Base We will convert all logarithms to base 3. Using the property $\log_{b^k} M = \frac{1}{k} \log_b M$, we rewrite $\log_9$ as $\log_{3^2}$. $2 \cdot \frac{1}{2} {\log _3}\,\,({3^{1 - x}} + 2) = 1 + {\log _3}\,\,({4.3^x} - 1)$ ${\log _3}\,\,({3^{1 - x}} + 2) = 1 + {\log _3}\,\,({4.3^x} - 1)$ Explanation: To combine and manipulate logarithmic terms effectively, they must share the same base. Base 3 is chosen because it is the base of the other given logarithm and $9$ is a power of $3$.

Step 3: Rearrange and Combine Logarithmic Terms We express the constant $1$ as $\log_3 3$. Then, we use the logarithm property $\log_b M + \log_b N = \log_b (MN)$ to combine the terms on the right side. ${\log _3}\,\,({3^{1 - x}} + 2) = {\log _3} 3 + {\log _3}\,\,({4.3^x} - 1)$ ${\log _3}\,\,({3^{1 - x}} + 2) = {\log _3}\,(3 \cdot ({4.3^x} - 1))$ Explanation: The goal here is to isolate the logarithmic terms and combine them into a single logarithm on each side of the equation. This simplifies the equation for the next step.

Step 4: Equate the Arguments of the Logarithms If $\log_b M = \log_b N$, then $M = N$. $3^{1 - x} + 2 = 3 \cdot (4.3^x - 1)$ Explanation: Once the equation is in the form of a single logarithm on each side with the same base, their arguments must be equal. This transforms the logarithmic equation into an algebraic one.

Step 5: Solve the Algebraic Equation Using Substitution First, simplify $3^{1-x} = \frac{3}{3^x}$. Let $y = 3^x$. Since $3^x > 0$ for all real $x$, we must have $y > 0$. $\frac{3}{y} + 2 = 3(4y - 1)$ $\frac{3 + 2y}{y} = 12y - 3$ Multiply both sides by $y$ (since $y \ne 0$): $3 + 2y = 12y^2 - 3y$ Rearrange into a standard quadratic equation: $12y^2 - 5y - 3 = 0$ Explanation: Equations involving expressions like $3^x$ and $3^{-x}$ are often simplified by substituting $y = 3^x$. This converts the equation into a familiar quadratic form, which can be solved using standard methods.

Step 6: Solve the Quadratic Equation for $y$ We factor the quadratic equation $12y^2 - 5y - 3 = 0$. We look for two numbers that multiply to $12 \times -3 = -36$ and add to $-5$. These numbers are $-9$ and $4$. $12y^2 - 9y + 4y - 3 = 0$ $3y(4y - 3) + 1(4y - 3) = 0$ $(3y + 1)(4y - 3) = 0$ This yields two possible values for $y$: $3y + 1 = 0 \implies y = -\frac{1}{3}$ $4y - 3 = 0 \implies y = \frac{3}{4}$ Explanation: Solving the quadratic equation gives the possible values for our substitution variable $y$.

Step 7: Substitute Back and Solve for $x$ Recall that $y = 3^x$.

• Case 1: $y = -\frac{1}{3}$ $3^x = -\frac{1}{3}$ This solution is not possible because $3^x$ is always positive for real values of $x$. Explanation: The range of the exponential function $f(x) = a^x$ (where $a > 0, a \ne 1$) is $(0, \infty)$. Thus, $3^x$ cannot be negative.

• Case 2: $y = \frac{3}{4}$ $3^x = \frac{3}{4}$ Take $\log_3$ on both sides to solve for $x$: $\log_3 (3^x) = \log_3 \left(\frac{3}{4}\right)$ $x = \log_3 3 - \log_3 4$ $x = 1 - \log_3 4$ Explanation: We use the definition of logarithm to solve for $x$. The property $\log_b b^k = k$ and the quotient rule for logarithms are applied here.

Step 8: Verify Domain Constraints We must ensure that the arguments of the original logarithms are positive for $x = 1 - \log_3 4$.

1. Argument of $\log_9$: $3^{1 - x} + 2$ Substitute $x = 1 - \log_3 4$: $3^{1 - (1 - \log_3 4)} + 2 = 3^{\log_3 4} + 2 = 4 + 2 = 6$. Since $6 > 0$, this is valid.

2. Argument of $\log_3$: $4 \cdot 3^x - 1$ Substitute $x = 1 - \log_3 4$: $4 \cdot 3^{(1 - \log_3 4)} - 1 = 4 \cdot 3^1 \cdot 3^{-\log_3 4} - 1 = 12 \cdot \frac{1}{3^{\log_3 4}} - 1 = 12 \cdot \frac{1}{4} - 1 = 3 - 1 = 2$. Since $2 > 0$, this is valid.

Both arguments are positive, so $x = 1 - \log_3 4$ is the correct solution. Explanation: This is a critical step in solving logarithmic equations. Any solution derived algebraically must be checked against the domain restrictions of the original logarithmic expressions to ensure it is not an extraneous solution.

Common Mistakes & Tips

• Domain Check is Crucial: Always verify that your solutions satisfy the domain requirements of the original logarithmic expressions. Extraneous solutions can arise from algebraic manipulations.
• Base Conversion: Ensure all logarithms are converted to a common base before attempting to combine them. This is a standard simplification technique.
• Substitution Strategy: When dealing with expressions like $a^x$ and $a^{-x}$, using a substitution (e.g., $y = a^x$) can transform the equation into a simpler polynomial form.

Summary The problem requires applying the definition of an arithmetic progression to three given terms involving logarithms. By converting all logarithms to a common base, using logarithm properties to simplify, and then solving the resulting algebraic equation (which involves a substitution), we find a potential value for $x$. Finally, it is essential to verify that this value of $x$ satisfies the domain constraints of the original logarithmic expressions to confirm it as the valid solution. The derived solution is $x = 1 - \log_3 4$.

The final answer is $\boxed{1 - \,{\log _3}\,4\,}$ which corresponds to option (B).