$l o g_{3} 2 l o g_{4} 3 l o g_{5} 4 . . . . . . l o g_{15} 14 l o g_{16} 15.$

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Solution

${log}_{3} 2 {log}_{4} 3 {log}_{5} 4 . . . . . . . . . . {log}_{15} 14 {log}_{16} 15 = ({log}_{3} 2 {log}_{4} 3) {log}_{5} 4 \times {log}_{6} 5 . . . . . . . . . . {log}_{15} 14 {log}_{16} 15 = {log}_{16} 2 = \frac{1}{{log}_{2} 16} = \frac{1}{{log}_{2} 2^{4}} = \frac{1}{4 {log}_{2} 2} = \frac{1}{4}$

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{log}_{3} 2 {log}_{4} 3 {log}_{5} 4..... {log}_{15} 14 {log}_{16} 15

l o g_{3} 2 l o g_{4} 3 l o g_{5} 4 \dots l o g_{15} 14 l o g_{16} 15

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