Match the following List I List II A. If $\alpha,\beta$ are the two roots of the equation$\displaystyle \dfrac{1-8(\log_{10}x)^2}{\log_{10}x-2(\log_{10}x^2)} =1$ then the value of $\displaystyle \frac{(\beta^3+1)}{\alpha^2\beta^3}$ is equal to 1. 5 B. Number of solution (s) of the equation $\log_2(x^2+3)=\dfrac{1}{2}\log_{1/3}\left(x+\dfrac{1}{x}\right),x >0$ is 2. 11 C. Integers satisfying the equation $|x|+\left|\dfrac{4-x^2}{x}\right|=\left|\dfrac{4}{x}\right|$ is\are 3. 0 D. If $\log_c2 \log_b 125=\log_{c}8$ where $c>0,c \neq1,b >1, b\neq$ 1, then $b$ is 4. 1

In part A, letting y = log10(x) and factoring the quadratic gives roots y = -1/4 and y = 1/2, leading to x values of 10^(-1/4) and 10^(1/2); substituting these into the expression yields 11, so A matches 2. For part B, simplifying the logarithmic equation to log2(x^2+3) = -1/2 * log3(x+1/x) shows that as x increases from 0 to infinity, the left side increases while the right side decreases, meaning their graphs intersect exactly once, so B matches 3. For part C, solving the absolute value equation across different intervals shows that there are no integer values of x that can satisfy the equation, meaning the answer is 0, so C matches 4. For part D, using the change of base formula, log_c(2) * log_b(125) = log_c(8) simplifies to log_c(2) * 3*log_b(5) = 3*log_c(2), which reduces to log_b(5) = 1 and gives b = 5, so D matches 1. The resulting matches are A-2, B-3, C-4, and D-1.

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