Find the value of \[\left( {{{\log }_3}11} \right)\left( {{{\log }_{11}}13} \right)\left( {{{\log }_{13}}15} \right)\left( {{{\log }_{15}}27} \right)\left( {{{\log }_{27}}81} \right)\].
Answer
383.1k+ views
Hint-Express ${\log _b}a = \dfrac{{\log a}}{{\log b}}$ and solve the problem.
So, by making use of this formula, now we can write the equation as
\[\left( {{{\log }_3}11} \right)\left( {{{\log }_{11}}13} \right)\left( {{{\log }_{13}}15} \right)\left( {{{\log }_{15}}27} \right)\left( {{{\log }_{27}}81} \right)\]
= $\left( {\dfrac{{\log 11}}{{\log 3}}} \right)\left( {\dfrac{{\log 13}}{{\log 11}}} \right)\left( {\dfrac{{\log 15}}{{\log 13}}} \right)\left( {\dfrac{{\log 27}}{{\log 15}}} \right)\left( {\dfrac{{\log 81}}{{\log 27}}} \right)$
So, from this equation log11, log13, log15, log27 will get cancelled out with each other
So, the remaining terms in the equation would now be
\[\dfrac{{\log 81}}{{\log 3}} = \dfrac{{\log {3^4}}}{{\log 3}}\]
To simplify this further, we will make use of the formula of $\log {a^m} = m\log a$
So, the equation can be further simplified and written as
=$\dfrac{{4\log 3}}{{\log 3}}$
So, from this the terms log3 will get cancelled out and we will have only 4 remaining
So, from this, we get \[\left( {{{\log }_3}11} \right)\left( {{{\log }_{11}}13} \right)\left( {{{\log }_{13}}15} \right)\left( {{{\log }_{15}}27} \right)\left( {{{\log }_{27}}81} \right)\]=4
Note: Make use of the appropriate logarithmic in accordance to the problem given, here we have made use of the formula ${\log _b}a = \dfrac{{\log a}}{{\log b}}$ primarily ,so in accordance to the problem given we have to make use of other logarithmic formulas too.
So, by making use of this formula, now we can write the equation as
\[\left( {{{\log }_3}11} \right)\left( {{{\log }_{11}}13} \right)\left( {{{\log }_{13}}15} \right)\left( {{{\log }_{15}}27} \right)\left( {{{\log }_{27}}81} \right)\]
= $\left( {\dfrac{{\log 11}}{{\log 3}}} \right)\left( {\dfrac{{\log 13}}{{\log 11}}} \right)\left( {\dfrac{{\log 15}}{{\log 13}}} \right)\left( {\dfrac{{\log 27}}{{\log 15}}} \right)\left( {\dfrac{{\log 81}}{{\log 27}}} \right)$
So, from this equation log11, log13, log15, log27 will get cancelled out with each other
So, the remaining terms in the equation would now be
\[\dfrac{{\log 81}}{{\log 3}} = \dfrac{{\log {3^4}}}{{\log 3}}\]
To simplify this further, we will make use of the formula of $\log {a^m} = m\log a$
So, the equation can be further simplified and written as
=$\dfrac{{4\log 3}}{{\log 3}}$
So, from this the terms log3 will get cancelled out and we will have only 4 remaining
So, from this, we get \[\left( {{{\log }_3}11} \right)\left( {{{\log }_{11}}13} \right)\left( {{{\log }_{13}}15} \right)\left( {{{\log }_{15}}27} \right)\left( {{{\log }_{27}}81} \right)\]=4
Note: Make use of the appropriate logarithmic in accordance to the problem given, here we have made use of the formula ${\log _b}a = \dfrac{{\log a}}{{\log b}}$ primarily ,so in accordance to the problem given we have to make use of other logarithmic formulas too.
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