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# Find the value of $\left( {{{\log }_3}11} \right)\left( {{{\log }_{11}}13} \right)\left( {{{\log }_{13}}15} \right)\left( {{{\log }_{15}}27} \right)\left( {{{\log }_{27}}81} \right)$.

Last updated date: 28th Mar 2023
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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.