If \[1,{\log _{10}}\left( {{4^x} - 2} \right),{\log _{10}}\left( {{4^x} + (\dfrac{{18}}{5})} \right)\] are in arithmetic progression for a real number \[x\], then the value of determinant \[\left| {\begin{array}{*{20}{c}}
  {2[x - \dfrac{1}{2}]}&{x - 1}&{{x^2}} \\
  1&0&x \\
  x&1&0
\end{array}} \right|\] is equal to

Hint : We are given that \[1,{\log _{10}}\left( {{4^x} - 2} \right),{\log _{10}}\left( {{4^x} + (\dfrac{{18}}{5})} \right)\] are in arithmetic progression (A.P) for a real number \[x\]. We have to find the value of determinant \[\left| {\begin{array}{*{20}{c}}
  {2[x - \dfrac{1}{2}]}&{x - 1}&{{x^2}} \\
  1&0&x \\
  x&1&0
\end{array}} \right|\].
We will be first finding the value of \[x\] by using the property of AP that in an AP sum of first and third terms gives twice the second term and use the below-mentioned formulae of logarithms. After finding the value of \[x\], we will be substituting that value in the determinant to find the value of the determinant.

Formula used : Following formulae will be useful for solving the given question.
If \[a,a + d,a + 2d\] are in AP then \[2(a + d) = a + a + 2d\]
\[
  b\log a = \log {(a)^b} \\
  1 + {\log _{10}}a = \log_{10} (10a) \\
 \]
Step-by-step Solution:
We know that in an AP sum of first and third terms gives twice the second term, i.e., If \[a,a + d,a + 2d\] are in AP then \[2(a + d) = a + a + 2d\]
So, by using this formula on the terms \[1,{\log _{10}}\left( {{4^x} - 2} \right),{\log _{10}}\left( {{4^x} + (\dfrac{{18}}{5})} \right)\], we get
\[2{\log _{10}}\left( {{4^x} - 2} \right) = 1 + {\log _{10}}\left( {{4^x} + (\dfrac{{18}}{5})} \right)\]
On using the logarithmic formulae \[b\log a = \log {(a)^b}\], \[1 + {\log _{10}}a = \log_{10}(10a)\], we get
\[
  {\log _{10}}{\left( {{4^x} - 2} \right)^2} = 1 + {\log _{10}}\left( {{4^x} + (\dfrac{{18}}{5})} \right) \\
  {\log _{10}}{\left( {{4^x} - 2} \right)^2} = {\log _{10}}10\left( {{4^x} + (\dfrac{{18}}{5})} \right) \\
 \]
We know that if \[\log a = \log b\] then \[a = b\].
So, on using this, we get
\[
  {\left( {{4^x} - 2} \right)^2} = 10\left( {{4^x} + (\dfrac{{18}}{5})} \right) \\
  {16^x} - 4 \times {4^x} + 4 = 10 \times {4^x} + (\dfrac{{18}}{5}) \times 10 \\
  {16^x} - 14 \times {4^x} + 4 - 36 = 0 \\
 \]
On further solving, we get
\[
  {16^x} - 14 \times {4^x} - 32 = 0 \\
  {16^x} + 2 \times {4^x} - 16 \times {4^x} - 32 = 0 \\
  {4^x}({4^x} + 2) - 16({4^x} + 2) = 0 \\
  ({4^x} - 16)({4^x} + 2) = 0 \\
 \]
On equating each term with \[0\], we get
\[
  {4^x} + 2 = 0 \\
  {4^x} = - 2 \\
 \]
Through this value of \[x\] cannot be found out.
Also,
\[
  {4^x} - 16 = 0 \\
  {4^x} = 16 \\
  {4^x} = {4^2} \\
 \]
On comparing, we get
\[x = 2\]
On substituting, this value in the determinant \[\left| {\begin{array}{*{20}{c}}
  {2[x - \dfrac{1}{2}]}&{x - 1}&{{x^2}} \\
  1&0&x \\
  x&1&0
\end{array}} \right|\], we get
\[
  \left| {\begin{array}{*{20}{c}}
  {2[2 - \dfrac{1}{2}]}&{2 - 1}&{{2^2}} \\
  1&0&2 \\
  2&1&0
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  {4 - 1}&1&4 \\
  1&0&2 \\
  2&1&0
\end{array}} \right| \\
   = \left| {\begin{array}{*{20}{c}}
  3&1&4 \\
  1&0&2 \\
  2&1&0
\end{array}} \right| \\
   = 3(0 - 2 \times 1) - 1(0 - 2 \times 2) + 4(1 \times 1 - 0) \\
   = 2 \\
 \]
Therefore the value of determinant is \[2\]

Note : Students generally make mistakes while using the property of Arithmetic Progression. Property of AP states that if \[a,a + d,a + 2d\] are in AP then \[2(a + d) = a + a + 2d\]. Students should this formula only to find the required values (here value of \[x\]).