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Question: Evaluate \(\dfrac{{{{\log }_3}135}}{{{{\log }_{15}}3}} - \dfrac{{{{\log }_3}5}}{{{{\log }_{405}}3}}\...

Evaluate log⁡3135log⁡153−log⁡35log⁡4053\dfrac{{{{\log }_3}135}}{{{{\log }_{15}}3}} - \dfrac{{{{\log }_3}5}}{{{{\log }_{405}}3}}

Explanation

Solution

In this sum the student has to use the properties of the logarithms which are log⁡(ab)=log⁡a+log⁡b\log (ab) = \log a + \log b,log⁡(ab)=log⁡a−log⁡b\log (\dfrac{a}{b}) = \log a - \log b,log⁡ab=log⁡blog⁡a{\log _a}b = \dfrac{{\log b}}{{\log a}}. The student has to use these properties one after the other. First step is using s. After that the student has to take the LCM and simplify the fraction before using the next property. In order to solve all the numericals related to logarithms the student needs to learn all the properties.

Complete step-by-step answer:

First step in evaluating logarithm numerical is to simplify them. Thus the given numerical can be simplified as
log⁡135×log⁡15log⁡3×log⁡3−log⁡5×log⁡405log⁡3×log⁡3........(1)\dfrac{{\log 135 \times \log 15}}{{\log 3 \times \log 3}} - \dfrac{{\log 5 \times \log 405}}{{\log 3 \times \log 3}}........(1)
Since the denominator of both the fractions are the same we can combine then and form a single fraction
log⁡135×log⁡15−log⁡5×log⁡405log⁡3×log⁡3..............(2)\dfrac{{\log 135 \times \log 15 - \log 5 \times \log 405}}{{\log 3 \times \log 3}}..............(2)
We can now make use of the property log⁡(ab)=log⁡a+log⁡b\log (ab) = \log a + \log band expand the numerator
log⁡135×(log⁡5+log⁡3)−log⁡5×(log⁡135+log⁡3)log⁡3×log⁡3..............(3)\dfrac{{\log 135 \times (\log 5 + \log 3) - \log 5 \times (\log 135 + \log 3)}}{{\log 3 \times \log 3}}..............(3)
log⁡135×log⁡5+log⁡135×log⁡3−log⁡5×log⁡135−log⁡5×log⁡3log⁡3×log⁡3..............(4)\dfrac{{\log 135 \times \log 5 + \log 135 \times \log 3 - \log 5 \times \log 135 - \log 5 \times \log 3}}{{\log 3 \times \log 3}}..............(4)
Removing log⁡135×log⁡5\log 135 \times \log 5as they are having opposite signs.
log⁡135×log⁡3−log⁡5×log⁡3log⁡3×log⁡3..............(5)\dfrac{{\log 135 \times \log 3 - \log 5 \times \log 3}}{{\log 3 \times \log 3}}..............(5)
It can be figured out from equation 33, that log⁡3\log 3is common in the numerator. We can take out log⁡3\log 3as a common factor and strike off one of the log⁡3\log 3 term from the denominator as well.
log⁡3×(log⁡135−log⁡5)log⁡3×log⁡3..............(6)\dfrac{{\log 3 \times (\log 135 - \log 5)}}{{\log 3 \times \log 3}}..............(6)
log⁡135−log⁡5log⁡3..............(7)\dfrac{{\log 135 - \log 5}}{{\log 3}}..............(7)
Now the last step is to make use of the propertylog⁡(ab)=log⁡a−log⁡b\log (\dfrac{a}{b}) = \log a - \log b.
⇒log⁡(1355)log⁡3..............(8)\Rightarrow \dfrac{{\log (\dfrac{{135}}{5})}}{{\log 3}}..............(8)
Simplifying the above equation to get the final answer.
⇒log⁡(27)log⁡3..............(9)\Rightarrow \dfrac{{\log (27)}}{{\log 3}}..............(9)
We can write log⁡27\log 27as 3log⁡33\log 3.
⇒3log⁡3log⁡3..............(9)\Rightarrow \dfrac{{3\log 3}}{{\log 3}}..............(9)

Thus the final answer for this sum is 33.

Note: Only way to solve these numericals is to learn the properties and then apply them step by step as and when necessary. Sometimes students may get confused with the properties and goof up while applying them. To prevent this from happening it is necessary that the student first notes down the property to be used in the sum in rough and then proceeds. Instead of blindly simplifying the sum , sometimes it is necessary to find a common factor and strike it off before simplifying. For example we did not simplify log⁡135\log 135 further, though it had common factors like 5,35,3. Instead we reduced log⁡405\log 405 in such a manner that we can bring it in the form of log⁡135\log 135 so that we could take out the common terms .This step made our sum much simpler.