Key Concepts and Formulas

• Arithmetic Progression (AP) Property: If three terms $a, b, c$ are in an arithmetic progression, then the middle term is the arithmetic mean of the other two: $2b = a + c$.
• Logarithm Domain Restriction: For $\log_B A$ to be defined, the argument $A$ must be positive ($A > 0$) and the base $B$ must be positive and not equal to 1 ($B > 0, B \neq 1$). In this problem, the base is 3.
• Logarithm Properties:
  • $k \log_b A = \log_b A^k$
  • $\log_b A + \log_b B = \log_b (AB)$
  • If $\log_b M = \log_b N$, then $M = N$.

Step-by-Step Solution

Step 1: Identify the terms and apply the AP property. The given terms are ${\log _3}2$, ${\log _3}({2^x} - 5)$, and ${\log _3}\left( {{2^x} - {7 \over 2}} \right)$. Since these terms are in an arithmetic progression, we can use the property $2b = a + c$. Let $a = {\log _3}2$, $b = {\log _3}({2^x} - 5)$, and $c = {\log _3}\left( {{2^x} - {7 \over 2}} \right)$. Applying the property, we get: $2{\log _3}({2^x} - 5) = {\log _3}2 + {\log _3}\left( {{2^x} - {7 \over 2}} \right)$

Step 2: Simplify both sides of the equation using logarithm properties. We use the property $k \log_b A = \log_b A^k$ on the left side: $2{\log _3}({2^x} - 5) = {\log _3}{({2^x} - 5)^2}$ We use the property $\log_b A + \log_b B = \log_b (AB)$ on the right side: ${\log _3}2 + {\log _3}\left( {{2^x} - {7 \over 2}} \right) = {\log _3}\left( {2 \cdot \left( {{2^x} - {7 \over 2}} \right)} \right)$ $= {\log _3}({2 \cdot 2^x} - 2 \cdot \frac{7}{2})$ $= {\log _3}({2^{x+1}} - 7)$

Step 3: Equate the arguments of the logarithms. Now, we have the equation: ${\log _3}{({2^x} - 5)^2} = {\log _3}({2^{x+1}} - 7)$ Since the logarithms have the same base, we can equate their arguments: ${({2^x} - 5)^2} = {2^{x+1}} - 7$

Step 4: Solve the resulting exponential equation by substitution. To simplify the equation, let $t = 2^x$. Then, $2^{x+1} = 2 \cdot 2^x = 2t$. Substitute $t$ into the equation: $(t - 5)^2 = 2t - 7$ Expand the left side: $t^2 - 10t + 25 = 2t - 7$ Rearrange the terms to form a quadratic equation: $t^2 - 10t - 2t + 25 + 7 = 0$ $t^2 - 12t + 32 = 0$

Step 5: Solve the quadratic equation for $t$. Factor the quadratic equation. We are looking for two numbers that multiply to 32 and add to -12. These numbers are -4 and -8. $(t - 4)(t - 8) = 0$ This gives two possible values for $t$: $t = 4 \quad \text{or} \quad t = 8$

Step 6: Substitute back to find the values of $x$. Recall that $t = 2^x$.

Case 1: $t = 4$ $2^x = 4$ $2^x = 2^2$ $x = 2$

Case 2: $t = 8$ $2^x = 8$ $2^x = 2^3$ $x = 3$

Step 7: Verify the solutions against the domain restrictions. For the logarithms to be defined, their arguments must be strictly positive.

1. ${\log _3}({2^x} - 5)$ requires $2^x - 5 > 0$, which means $2^x > 5$.
2. ${\log _3}\left( {{2^x} - {7 \over 2}} \right)$ requires $2^x - \frac{7}{2} > 0$, which means $2^x > \frac{7}{2}$ or $2^x > 3.5$.

Let's check our potential solutions:

For $x = 2$: $2^x = 2^2 = 4$. Check $2^x > 5$: Is $4 > 5$? No. Since the condition $2^x > 5$ is not met, $x=2$ is an extraneous solution.

For $x = 3$: $2^x = 2^3 = 8$. Check $2^x > 5$: Is $8 > 5$? Yes. Check $2^x > 3.5$: Is $8 > 3.5$? Yes. Since both conditions are met, $x=3$ is a valid solution.

Common Mistakes & Tips

• Ignoring Domain Restrictions: Always check the arguments of logarithms to ensure they are positive. Failing to do so can lead to extraneous solutions.
• Incorrect Logarithm Properties: Double-check the application of properties like $k \log A = \log A^k$ and $\log A + \log B = \log(AB)$.
• Algebraic Errors: Be careful when expanding squares and rearranging terms in quadratic equations.

Summary

The problem requires combining the property of arithmetic progressions with the properties and domain restrictions of logarithms. By setting up the equation using the AP property, simplifying with logarithm rules, and solving the resulting algebraic equation, we obtained two potential values for $x$. Crucially, we verified these values against the domain requirements for the original logarithmic expressions, identifying that only one of them is a valid solution.

The final answer is $\boxed{3}$.