Question

Let $x<1$, then value of $\begin{vmatrix} x^2+2 & 2x+1 & 2log_4 2 \\ 2x+1 & x+2 & 2log_9 3 \\ log_2 8 & log_3 27 & 2log_{16} 4 \end{vmatrix}$

Answer 1

$(x-1)^2(x-3)$

Answer 2

To evaluate the given determinant, we first simplify the logarithmic terms.

The determinant is: $D = \begin{vmatrix} x^2+2 & 2x+1 & 2log_4 2 \\ 2x+1 & x+2 & 2log_9 3 \\ log_2 8 & log_3 27 & 2log_{16} 4 \end{vmatrix}$

Let's simplify each logarithmic term:

1. $2\log_4 2 = \log_4 (2^2) = \log_4 4 = 1$ (Alternatively, using change of base formula: $2 \times \frac{\log_2 2}{\log_2 4} = 2 \times \frac{1}{2} = 1$)

2. $2\log_9 3 = \log_9 (3^2) = \log_9 9 = 1$ (Alternatively: $2 \times \frac{\log_3 3}{\log_3 9} = 2 \times \frac{1}{2} = 1$)

3. $\log_2 8 = \log_2 (2^3) = 3$

4. $\log_3 27 = \log_3 (3^3) = 3$

5. $2\log_{16} 4 = \log_{16} (4^2) = \log_{16} 16 = 1$ (Alternatively: $2 \times \frac{\log_4 4}{\log_4 16} = 2 \times \frac{1}{2} = 1$)

Substitute these simplified values back into the determinant: $D = \begin{vmatrix} x^2+2 & 2x+1 & 1 \\ 2x+1 & x+2 & 1 \\ 3 & 3 & 1 \end{vmatrix}$

Now, we can simplify the determinant using row operations. Apply the operations $R_1 \to R_1 - R_3$ and $R_2 \to R_2 - R_3$:

For $R_1$: $(x^2+2) - 3 = x^2 - 1$ $(2x+1) - 3 = 2x - 2$ $1 - 1 = 0$

For $R_2$: $(2x+1) - 3 = 2x - 2$ $(x+2) - 3 = x - 1$ $1 - 1 = 0$

The determinant becomes: $D = \begin{vmatrix} x^2-1 & 2x-2 & 0 \\ 2x-2 & x-1 & 0 \\ 3 & 3 & 1 \end{vmatrix}$

Expand the determinant along the third column ($C_3$): $D = 0 \cdot (\text{cofactor of } (1,3)) - 0 \cdot (\text{cofactor of } (2,3)) + 1 \cdot \begin{vmatrix} x^2-1 & 2x-2 \\ 2x-2 & x-1 \end{vmatrix}$ $D = 1 \cdot [(x^2-1)(x-1) - (2x-2)(2x-2)]$

Factor out common terms: $x^2-1 = (x-1)(x+1)$ $2x-2 = 2(x-1)$

Substitute these into the expression for $D$: $D = (x-1)(x+1)(x-1) - [2(x-1)][2(x-1)]$ $D = (x-1)^2(x+1) - 4(x-1)^2$

Now, factor out $(x-1)^2$: $D = (x-1)^2 [(x+1) - 4]$ $D = (x-1)^2 (x-3)$

The given condition $x<1$ does not further simplify the expression, it just specifies the domain for $x$.

The final answer is $(x-1)^2(x-3)$.

Explanation of the solution:

1. Simplify all logarithmic terms in the determinant using logarithm properties.

2. Substitute the numerical values of the simplified logarithmic terms back into the determinant.

3. Apply row operations ($R_1 \to R_1 - R_3$ and $R_2 \to R_2 - R_3$) to create zeros in the third column, simplifying the expansion.

4. Expand the determinant along the third column.

5. Factorize the terms in the resulting $2 \times 2$ determinant and simplify the expression to obtain the final value.