The solution set of the equation left| {1 - { | vedclass

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(A) Let $u = \log_6 x$. Then $\log_{1/6} x = -u$,$\log_2 x = \log_2 6 \cdot u$,and $\log_{1/2} x = -\log_2 6 \cdot u$.Given equation: $|1 + u| + |u \l

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$5$

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(A) Let $u = \log_6 x$. Then $\log_{1/6} x = -u$,$\log_2 x = \log_2 6 \cdot u$,and $\log_{1/2} x = -\log_2 6 \cdot u$.Given equation: $|1 + u| + |u \log_2 6| + 2 = |3 + u - u \log_2 6|$.Using the property $|A| + |B| + |C| = |A+B+C|$ if $A, B, C$ have the same sign is not directly applicable,but we note $|x| + |y| + |z| = |x+y+z|$ holds if $x, y, z$ are of the same sign.Here,let $A = 1+u$,$B = u \log_2 6$,$C = 1$. Then $|A| + |B| + |2| = |A+B+2|$ implies $A, B, 2$ have the same sign.Since $2 > 0$,we must have $1+u \ge 0$ and $u \log_2 6 \ge 0$,which implies $u \ge 0$.Thus $\log_6 x \ge 0 \implies x \ge 1$.Also,the equation simplifies to $1+u + u \log_2 6 + 2 = 3 + u - u \log_2 6$ is not correct; rather,the condition $|A| + |B| + |C| = |A+B+C|$ implies $A, B, C$ have the same sign.Solving the inequality $1+u \ge 0$ and $u \log_2 6 \ge 0$ gives $u \ge 0$,so $x \ge 1$.For the equality to hold,we need $u \log_2 6 \le 2$ (from the structure of the equation).$u \le \frac{2}{\log_2 6} = 2 \log_6 2 = \log_6 4$.So $0 \le \log_6 x \le \log_6 4$,which means $1 \le x \le 4$.Thus the interval is $[1, 4]$,so $a=4, b=1$. $a+b = 5$.

The solution set of the equation $\left| {1 - {{\log }_{1/6}}x} \right| + \left| {{{\log }_2}x} \right| + 2 = \left| {3 - {{\log }_{1/6}}x + {{\log }_{1/2}}x} \right|$ is $\left[ {\frac{a}{b},a} \right]$,where $a, b \in N$. Then the value of $(a + b)$ is:

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