Question

Find the value of the expression ${{\log }_{3}}4{{\log }_{4}}5{{\log }_{5}}6{{\log }_{6}}7{{\log }_{7}}8{{\log }_{8}}9$

Answer 1

Hint: Use the base changing formula of the logarithm, i.e. ${{\log }_{b}}a=\dfrac{{{\log }_{c}}a}{{{\log }_{c}}b}$. Hence convert the base of all the logarithms involved to a common base (say 10) and hence find the value of the given expression.

Complete step-by-step answer:
Using base changing formula to convert the base of ${{\log }_{3}}4$ to e, we get
${{\log }_{3}}4=\dfrac{\ln 4}{\ln 3}\text{ }\left( i \right)$
Using base changing formula to convert the base of ${{\log }_{4}}5$ to e, we get
${{\log }_{4}}5=\dfrac{\ln 5}{\ln 4}\text{ }\left( ii \right)$
Using base changing formula to convert the base of ${{\log }_{5}}6$ to e, we get
${{\log }_{5}}6=\dfrac{\ln 6}{\ln 5}\text{ }\left( iii \right)$
Using base changing formula to convert the base of ${{\log }_{6}}7$ to e, we get
${{\log }_{6}}7=\dfrac{\ln 7}{\ln 6}\text{ }\left( iv \right)$
Using base changing formula to convert the base of ${{\log }_{7}}8$ to e, we get
${{\log }_{7}}8=\dfrac{\ln 8}{\ln 7}\text{ }\left( v \right)$
Using base changing formula to convert the base of ${{\log }_{8}}9$ to e, we get
${{\log }_{8}}9=\dfrac{\ln 9}{\ln 8}\text{ }\left( vi \right)$
Multiplying equation(i), equation (ii), equation (iii), equation (iv), equation (v) and equation (vi), we get
${{\log }_{3}}4{{\log }_{4}}5{{\log }_{5}}6{{\log }_{6}}7{{\log }_{7}}8{{\log }_{8}}9=\dfrac{\ln 4}{\ln 3}\times \dfrac{\ln 5}{\ln 4}\times \dfrac{\ln 6}{\ln 5}\times \dfrac{\ln 7}{\ln 6}\times \dfrac{\ln 8}{\ln 7}\times \dfrac{\ln 9}{\ln 8}$
Simplifying, we get
${{\log }_{3}}4{{\log }_{4}}5{{\log }_{5}}6{{\log }_{6}}7{{\log }_{7}}8{{\log }_{8}}9=\dfrac{\ln 9}{\ln 3}$
We know that $\ln {{m}^{n}}=n\ln m$
Hence, we have
${{\log }_{3}}4{{\log }_{4}}5{{\log }_{5}}6{{\log }_{6}}7{{\log }_{7}}8{{\log }_{8}}9=\dfrac{\ln 9}{\ln 3}=\dfrac{\ln {{3}^{2}}}{\ln 3}=\dfrac{2\ln 3}{\ln 3}=2$
Hence the value of the given expression is 2.

Note: Alternative solution:
We know from base changing formula
${{\log }_{a}}c=\dfrac{{{\log }_{b}}c}{{{\log }_{b}}a}$
Multiplying both sides by ${{\log }_{b}}a$, we get
${{\log }_{b}}c={{\log }_{b}}a{{\log }_{a}}c$
Hence, we have
${{\log }_{3}}4{{\log }_{4}}5={{\log }_{3}}5$
Multiplying both sides by ${{\log }_{5}}6$, we get
${{\log }_{3}}4{{\log }_{4}}5{{\log }_{5}}6={{\log }_{3}}5{{\log }_{5}}6={{\log }_{3}}6$
Continuing in this way, we get
${{\log }_{3}}4{{\log }_{4}}5{{\log }_{5}}6{{\log }_{6}}7{{\log }_{7}}8{{\log }_{8}}9={{\log }_{3}}9={{\log }_{3}}{{3}^{2}}$
We know that ${{\log }_{a}}{{a}^{n}}=n$
Hence, we get
${{\log }_{3}}4{{\log }_{4}}5{{\log }_{5}}6{{\log }_{6}}7{{\log }_{7}}8{{\log }_{8}}9=2$, which is the same as obtained above.