Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

How do you find the logarithm \[{\log _{\dfrac{2}{3}}}\left( {\dfrac{8}{{27}}} \right)\] ?

Answer
VerifiedVerified
517.5k+ views
Hint: In the given question, we have been asked to find the logarithm of \[{\log _{\dfrac{2}{3}}}\left( {\dfrac{8}{{27}}} \right)\] . To calculate this value, we need to keep two formulas in our minds. The first one is \[{\log _{_e}}{X^n} = n{\log _e}X\] and the second formula is \[{\log _{_e}}e = 1\] . With the help of these two formulas, we will now evaluate the value of the above question.

Formula used:
The first formula that we will use is: \[{\log _{_e}}{X^n} = n{\log _e}X\].
The second formula that we should keep in mind is: \[{\log _{_e}}e = 1\].

Complete step by step solution:
The logarithm given to us is \[{\log _{\dfrac{2}{3}}}\left( {\dfrac{8}{{27}}} \right)\].
This can also be written as \[{\log _{\dfrac{2}{3}}}\left( {\dfrac{{{2^3}}}{{{3^3}}}} \right)\].
We can see that the power of both \[2\]and \[3\]is equal. Therefore, we can change this to \[{\log _{\dfrac{2}{3}}}{\left( {\dfrac{2}{3}} \right)^3}\].
Now in the next step, we will use the formula 1), that is, \[{\log _{_e}}{X^n} = n{\log _e}X\]. The value of n will be replaced by 3, on doing this we get: \[\left( 3 \right){\log _{\dfrac{2}{3}}}\left( {\dfrac{2}{3}} \right)\] .
In the next step, we will use the formula 2) \[{\log _{_e}}e = 1\], here we will replace the value of \[e\] with \[\dfrac{2}{3}\].
 This will change the equation to:
\[3\left( 1 \right)\]
\[ \Rightarrow 3\] .
From this we can derive that the value of logarithm \[{\log _{\dfrac{2}{3}}}\left( {\dfrac{8}{{27}}} \right)\] is \[3\] .

Note: In order to solve these types of questions involving logarithm, you should always need to remember the formulas of logarithm. For example, we used two formulas of logarithm in the above solution: 1) \[{\log _{_e}}{X^n} = n{\log _e}X\] and 2) \[{\log _{_e}}e = 1\]. These formulas will help you in quickly solving the solutions and they will also make the question easier to solve.