Question

How do you evaluate ${\log _{81}}27$?

Answer 1

Simply implement 27 and 81 into ${x^y}$form and implement log properties to get the answer Here, first what you have to do is that you have to replace 27 with ${3^3}$and 81 with ${3^4}$.
After which we will use a special log formula which states that ${\log _x}y = \dfrac{{\log y}}{{\log x}}$.
After modifying the whole equation we will use the formula of ${\log _x}{y^z} = z{\log _x}y$ to modify both the numerator and denominator. Solving then, we will cancel out the like terms which will be of the form ${\log _x}y$to get a simple numeral fraction of the form $\dfrac{x}{y}$to get the answer.

Complete step by step solution:
Here, the given problem is to solve ${\log _{81}}27$
we know that,
$27 = {3^3}$
And
$81 = {3^4}$
Therefore, we can rewrite the given question as
${\log _{81}}27 = {\log _{{3^4}}}{3^3}$
here, we will use a special property of log which states that,
${\log _x}y = \dfrac{{\log y}}{{\log x}}$
using the above property we will rewrite our equation as
${\log _{{3^4}}}{3^3} = \dfrac{{\log {3^3}}}{{\log {3^4}}} \\\ \Rightarrow \dfrac{{3\log 3}}{{4\log 3}} = \dfrac{3}{4} \\\$
Hence, value of ${\log _{81}}27$is $\dfrac{3}{4}$

Note: The property, ${\log _x}y = \dfrac{{\log y}}{{\log x}}$ actually states that if you have a log function like ${\log _y}x$, then you can write the same function as $\dfrac{{{{\log }_z}x}}{{{{\log }_z}y}}$where the bases of log must be the same. You can take the base as anything, it can be 5,10,15,10000 etc.
You should always remember to make use of this property such that the base is consistent with the question. Since these questions didn’t need any specific requirement so we just took the normal base 10 log as it is. But you can also take a log with base 100 or any other number to solve this question.