Suppose that 4^(x_1) = 5, 5^(x_2) = 6, 6^(x_3) = 7,...., 126^(x_123) = 127, 127^(x_124) = 128. What is the value of the product x_1x_2...x_124?

Nov 27, 2016

$3 \frac{1}{2}$

Explanation:

${4}^{{x}_{1}} = 5$. Taking log of both sides we get ${x}_{1} \log 4 = \log 5 \mathmr{and} {x}_{1} = \log \frac{5}{\log} 4$.
${5}^{{x}_{2}} = 6$. Taking log of both sides we get ${x}_{2} \log 5 = \log 6 \mathmr{and} {x}_{2} = \log \frac{6}{\log} 5$.
${6}^{{x}_{3}} = 7$. Taking log of both sides we get ${x}_{1} \log 6 = \log 7 \mathmr{and} {x}_{3} = \log \frac{7}{\log} 6$.
$\ldots \ldots \ldots \ldots \ldots \ldots$
${126}^{{x}_{123}} = 127$. Taking log of both sides we get ${x}_{123} \log 126 = \log 127 \mathmr{and} {x}_{123} = \log \frac{127}{\log} 126$.
${127}^{{x}_{124}} = 128$. Taking log of both sides we get ${x}_{124} \log 127 = \log 128 \mathmr{and} {x}_{124} = \log \frac{128}{\log} 127$.
${x}_{1} \cdot {x}_{2} \cdot \ldots . \cdot x 124 = \left(\cancel{\log} \frac{5}{\log} 4\right) \left(\cancel{\log} \frac{6}{\cancel{\log}} 5\right) \left(\cancel{\log} \frac{7}{\cancel{\log}} 6\right) \ldots \log \left(\frac{\cancel{127}}{\cancel{\log}} 126\right) \left(\log \frac{128}{\cancel{\log}} 127\right) = \log \frac{128}{\log} 4 = \log {2}^{7} / \log {2}^{2} = \frac{7 \cancel{\log} 2}{2 \cancel{\log} 2} = \frac{7}{2} = 3 \frac{1}{2}$[Ans]