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Step-by-Step Solution
Step 1: Understand the Arithmetic Progression condition
We are given that the three terms 1, log_{10}(4^x - 2) , and log_{10}\bigl(4^x + \frac{18}{5}\bigr) are in arithmetic progression (AP).
By definition of AP, if three terms a, b, c are in AP, then:
2b = a + c .
Here,
a = 1, \;
b = log_{10}(4^x - 2), \;
c = log_{10}\bigl(4^x + \frac{18}{5}\bigr).
So the AP condition becomes:
2 \, log_{10}(4^x - 2) = 1 + log_{10}\bigl(4^x + \frac{18}{5}\bigr).
Step 2: Simplify the AP relation using logarithm properties
Using the property of logarithms,
2 \, log_{10}(4^x - 2) = log_{10}\bigl((4^x - 2)^2\bigr) .
Hence the given equation becomes:
log_{10}\bigl((4^x - 2)^2\bigr) = 1 + log_{10}\bigl(4^x + \frac{18}{5}\bigr).
Recall that 1 in the logarithm base 10 means log_{10}(10) . Therefore:
log_{10}\bigl((4^x - 2)^2\bigr) = log_{10}(10) + log_{10}\bigl(4^x + \frac{18}{5}\bigr).
By the log property
log_{10}(A) + log_{10}(B) = log_{10}(AB) ,
we get:
log_{10}\bigl((4^x - 2)^2\bigr) = log_{10}\Bigl(10 \times \bigl(4^x + \frac{18}{5}\bigr)\Bigr).
Step 3: Equate the inside expressions of the logarithms
Since log_{10}(A) = log_{10}(B) implies A = B for positive A and B, we have:
(4^x - 2)^2 = 10 \left(4^x + \frac{18}{5}\right).
Step 4: Expand and simplify the quadratic equation in 4^x
Expand the left-hand side:
(4^x - 2)^2 = (4^x)^2 - 2 \cdot 4^x \cdot 2 + 2^2 = (4^x)^2 - 4 \cdot 4^x + 4.
The right-hand side is:
10 \left(4^x + \frac{18}{5}\right) = 10 \cdot 4^x + 10 \cdot \frac{18}{5} = 10 \cdot 4^x + 36.
So the equation becomes:
(4^x)^2 - 4 \cdot 4^x + 4 = 10 \cdot 4^x + 36.
Rearrange terms:
(4^x)^2 - 4 \cdot 4^x - 10 \cdot 4^x + 4 - 36 = 0, \\
(4^x)^2 - 14 \cdot 4^x - 32 = 0.
Step 5: Factorize the equation in terms of 4^x
Factor out common terms:
(4^x)^2 - 16 \cdot 4^x + 2 \cdot 4^x - 32 = 0, \\
(4^x)(4^x - 16) + 2(4^x - 16) = 0, \\
(4^x + 2)(4^x - 16) = 0.
So the solutions for 4^x are -2 or 16. Since 4^x must be positive for any real x, 4^x = -2 is not possible.
Thus, 4^x = 16 .
Hence, 4^x = 4^2 \; \Rightarrow \; x = 2.
Step 6: Substitute x = 2 into the given determinant
We want to find the value of the determinant
\left|
\begin{matrix}
2\left(x - \frac{1}{2}\right) & x - 1 & x^2 \\
1 & 0 & x \\
x & 1 & 0
\end{matrix}
\right|.
Substitute x = 2 :
1. 2\left(2 - \frac{1}{2}\right) = 2 \times \frac{3}{2} = 3.
2. 2 - 1 = 1.
3. (2)^2 = 4.
So the determinant becomes:
\left|
\begin{matrix}
3 & 1 & 4 \\
1 & 0 & 2 \\
2 & 1 & 0
\end{matrix}
\right|.
Step 7: Compute the 3×3 determinant
Recall the expansion of a 3×3 determinant along the first row:
\begin{vmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{vmatrix}
= a \,(ei - fh) - b \,(di - fg) + c \,(dh - eg).
Using this for our matrix:
\begin{vmatrix}
3 & 1 & 4 \\
1 & 0 & 2 \\
2 & 1 & 0
\end{vmatrix}
= 3 \times \bigl((0)(0) - (2)(1)\bigr)
- 1 \times \bigl((1)(0) - (2)(2)\bigr)
+ 4 \times \bigl((1)(1) - (0)(2)\bigr).
Simplify each term:
1. 3 \times (0 - 2) = 3 \times (-2) = -6.
2. -1 \times (0 - 4) = -1 \times (-4) = 4.
3. 4 \times (1 - 0) = 4 \times 1 = 4.
Add them up:
-6 + 4 + 4 = 2.
Final Answer
The value of the determinant is 2 .
