Let p = 99 and q = 101. Define p_1 = log_{10} | vedclass

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(A) Given $p = 99$ and $q = 101$. $p_1 = \log_{10} \left(\frac{99+101}{2}\right) = \log_{10} 100 = 2$. $q_1 = \frac{1}{2}(\log_{10} 99 + \log_{10} 101

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$\log p_1 > p_2 > q_2 > \log q_1$

Vedclass · Answer

(A) Given $p = 99$ and $q = 101$. $p_1 = \log_{10} \left(\frac{99+101}{2}ight) = \log_{10} 100 = 2$. $q_1 = \frac{1}{2}(\log_{10} 99 + \log_{10} 101) = \log_{10} \sqrt{99 	imes 101} = \log_{10} \sqrt{9999}$. Since $9999 Thus,$p_1 > q_1$. By the Arithmetic Mean-Geometric Mean inequality,for any two positive numbers $a$ and $b$,$\frac{a+b}{2} > \sqrt{ab}$ if $a 
eq b$. Since $p_1 > q_1$,we have $\frac{p_1+q_1}{2} > \sqrt{p_1 q_1}$. Taking $\log_{10}$ on both sides,$\log_{10} \left(\frac{p_1+q_1}{2}ight) > \log_{10} \sqrt{p_1 q_1} = \frac{1}{2}(\log_{10} p_1 + \log_{10} q_1)$. This implies $p_2 > q_2$. Also,since $p_1 > q_1$,it follows that $\log_{10} p_1 > \log_{10} q_1$. Since $p_1 > \frac{p_1+q_1}{2} > q_1$,applying the logarithm function (which is strictly increasing) gives $\log_{10} p_1 > p_2 > \log_{10} q_1$. Combining these,we get $\log_{10} p_1 > p_2 > q_2 > \log_{10} q_1$.

Let $p = 99$ and $q = 101$. Define $p_1 = \log_{10} \left(\frac{p+q}{2}\right)$ and $q_1 = \frac{1}{2}(\log_{10} p + \log_{10} q)$,and $p_2 = \log_{10} \left(\frac{p_1+q_1}{2}\right)$,$q_2 = \frac{1}{2}(\log_{10} p_1 + \log_{10} q_1)$. Then:

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