Key Concepts and Formulas

Before we dive into the solution, let's recall the fundamental concepts we'll be using:

1. Arithmetic Progression (AP): Three numbers $a, b, c$ are in AP if the middle term is the average of the other two, i.e., $2b = a+c$.
2. Logarithm Properties:
   • Domain: For $\log_b M$ to be defined, $M > 0$ and $b > 0, b \neq 1$.
   • Product Rule: $\log_b M + \log_b N = \log_b (MN)$
   • Power Rule: $k \log_b M = \log_b (M^k)$
   • Base Identity: $\log_b b = 1$
   • Equality: If $\log_b M = \log_b N$, then $M=N$.
3. Determinant of a $3 \times 3$ Matrix: For a matrix $A = \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}$, its determinant is given by $a(ei - fh) - b(di - fg) + c(dh - eg)$.

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Step-by-Step Solution

Part 1: Finding the value of $x$

We are given that $1$, $\log_{10} (4^x - 2)$, and $\log_{10} \left( 4^x + \frac{18}{5} \right)$ are in arithmetic progression.

1. Establish the conditions for logarithms to be defined: For the logarithmic terms to be real and defined, their arguments must be positive:

• $4^x - 2 > 0 \implies 4^x > 2$
• $4^x + \frac{18}{5} > 0$ (This condition is always satisfied since $4^x$ is always positive for real $x$, and $\frac{18}{5}$ is positive). So, the crucial condition is $4^x > 2$.

2. Apply the AP condition: If $a, b, c$ are in AP, then $2b = a+c$. Here, $a=1$, $b=\log_{10} (4^x - 2)$, and $c=\log_{10} \left( 4^x + \frac{18}{5} \right)$. Substituting these values: $2 \log_{10} (4^x - 2) = 1 + \log_{10} \left( 4^x + \frac{18}{5} \right)$

3. Simplify the logarithmic equation using properties:

• First, use the power rule $k \log_b M = \log_b (M^k)$ on the left side: $\log_{10} (4^x - 2)^2 = 1 + \log_{10} \left( 4^x + \frac{18}{5} \right)$
• Next, convert the constant term $1$ into a logarithm with base 10 using $\log_b b = 1$: $\log_{10} (4^x - 2)^2 = \log_{10} 10 + \log_{10} \left( 4^x + \frac{18}{5} \right)$
• Now, use the product rule $\log_b M + \log_b N = \log_b (MN)$ on the right side: $\log_{10} (4^x - 2)^2 = \log_{10} \left( 10 \cdot \left( 4^x + \frac{18}{5} \right) \right)$

4. Equate the arguments of the logarithms: Since the logarithms on both sides have the same base, their arguments must be equal: $(4^x - 2)^2 = 10 \left( 4^x + \frac{18}{5} \right)$

5. Solve the resulting exponential equation:

• Expand the left side and distribute on the right side: $(4^x)^2 - 2 \cdot 2 \cdot 4^x + 2^2 = 10 \cdot 4^x + 10 \cdot \frac{18}{5}$ $(4^x)^2 - 4 \cdot 4^x + 4 = 10 \cdot 4^x + 36$
• Rearrange the terms to form a quadratic equation in terms of $4^x$: $(4^x)^2 - 4 \cdot 4^x - 10 \cdot 4^x + 4 - 36 = 0$ $(4^x)^2 - 14 \cdot 4^x - 32 = 0$
• Let $y = 4^x$. The equation becomes a quadratic equation in $y$: $y^2 - 14y - 32 = 0$
• Factor the quadratic equation: We need two numbers that multiply to -32 and add to -14. These are -16 and 2. $y^2 - 16y + 2y - 32 = 0$ $y(y - 16) + 2(y - 16) = 0$ $(y + 2)(y - 16) = 0$
• Substitute back $y = 4^x$: $(4^x + 2)(4^x - 16) = 0$
• This gives two possible solutions for $4^x$: $4^x + 2 = 0 \implies 4^x = -2$ $4^x - 16 = 0 \implies 4^x = 16$

6. Validate the solutions for $x$:

• For $4^x = -2$: An exponential term $a^x$ (where $a > 0$) is always positive for any real $x$. Therefore, $4^x = -2$ has no real solution for $x$. This solution is extraneous.
• For $4^x = 16$: We can write $16$ as $4^2$. $4^x = 4^2$ $x = 2$
• Finally, check this value of $x$ against the domain condition $4^x > 2$: For $x=2$, $4^2 = 16$. Since $16 > 2$, the value $x=2$ is valid.

So, the only valid value for $x$ is $2$.

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Part 2: Evaluating the Determinant

Now we need to calculate the value of the given determinant with $x=2$.

1. Substitute $x=2$ into the determinant: The given determinant is: $\begin{vmatrix} 2\left( x - \frac{1}{2} \right) & x - 1 & x^2 \\ 1 & 0 & x \\ x & 1 & 0 \end{vmatrix}$ Substitute $x=2$:

• $2\left( x - \frac{1}{2} \right) = 2\left( 2 - \frac{1}{2} \right) = 2\left( \frac{3}{2} \right) = 3$
• $x - 1 = 2 - 1 = 1$
• $x^2 = 2^2 = 4$
• The elements $1, 0, x$ become $1, 0, 2$.
• The elements $x, 1, 0$ become $2, 1, 0$.

The determinant becomes: $\begin{vmatrix} 3 & 1 & 4 \\ 1 & 0 & 2 \\ 2 & 1 & 0 \end{vmatrix}$

2. Evaluate the $3 \times 3$ determinant: Using cofactor expansion along the first row: $\text{Determinant} = 3 \cdot \begin{vmatrix} 0 & 2 \\ 1 & 0 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 2 \\ 2 & 0 \end{vmatrix} + 4 \cdot \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix}$

• Calculate the $2 \times 2$ determinants:
  • $\begin{vmatrix} 0 & 2 \\ 1 & 0 \end{vmatrix} = (0 \cdot 0) - (2 \cdot 1) = 0 - 2 = -2$
  • $\begin{vmatrix} 1 & 2 \\ 2 & 0 \end{vmatrix} = (1 \cdot 0) - (2 \cdot 2) = 0 - 4 = -4$
  • $\begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} = (1 \cdot 1) - (0 \cdot 2) = 1 - 0 = 1$
• Substitute these values back: $\text{Determinant} = 3(-2) - 1(-4) + 4(1)$ $\text{Determinant} = -6 + 4 + 4$ $\text{Determinant} = -6 + 8$ $\text{Determinant} = 2$

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Summary and Key Takeaways

This problem beautifully combines concepts from Arithmetic Progressions, Logarithms, Exponential Equations, and Determinants. The key steps involved:

1. Setting up the AP relation: Correctly translating the "in AP" condition into an equation.
2. Logarithm Manipulation: Applying logarithm properties to simplify the equation, including converting constants to logarithmic form and using product/power rules.
3. Domain Consideration: Always checking the domain of logarithmic functions (argument must be positive) to filter out extraneous solutions. This is a common pitfall.
4. Solving Exponential/Quadratic Equations: Recognizing and solving the quadratic equation formed by substitution ($y = 4^x$).
5. Determinant Evaluation: Accurately substituting the obtained value of $x$ and then expanding the $3 \times 3$ determinant.

Always remember to check for domain restrictions, especially with logarithms and square roots, as they often lead to invalid solutions.