Evaluate $\dfrac{{{{\log }_3}135}}{{{{\log }_{15}}3}} - \dfrac{{{{\log }_3}5}}{{{{\log }_{405}}3}}$
Answer
585.6k+ views
Hint: In this sum the student has to use the properties of the logarithms which are $\log (ab) = \log a + \log b$,$\log (\dfrac{a}{b}) = \log a - \log b$,${\log _a}b = \dfrac{{\log b}}{{\log a}}$. 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
$\dfrac{{\log 135 \times \log 15}}{{\log 3 \times \log 3}} - \dfrac{{\log 5 \times \log 405}}{{\log 3 \times \log 3}}........(1)$
Since the denominator of both the fractions are the same we can combine then and form a single fraction
$\dfrac{{\log 135 \times \log 15 - \log 5 \times \log 405}}{{\log 3 \times \log 3}}..............(2)$
We can now make use of the property $\log (ab) = \log a + \log b$and expand the numerator
$\dfrac{{\log 135 \times (\log 5 + \log 3) - \log 5 \times (\log 135 + \log 3)}}{{\log 3 \times \log 3}}..............(3)$
$\dfrac{{\log 135 \times \log 5 + \log 135 \times \log 3 - \log 5 \times \log 135 - \log 5 \times \log 3}}{{\log 3 \times \log 3}}..............(4)$
Removing $\log 135 \times \log 5$as they are having opposite signs.
$\dfrac{{\log 135 \times \log 3 - \log 5 \times \log 3}}{{\log 3 \times \log 3}}..............(5)$
It can be figured out from equation $3$, that $\log 3$is common in the numerator. We can take out $\log 3$as a common factor and strike off one of the $\log 3$ term from the denominator as well.
\[\dfrac{{\log 3 \times (\log 135 - \log 5)}}{{\log 3 \times \log 3}}..............(6)\]
$\dfrac{{\log 135 - \log 5}}{{\log 3}}..............(7)$
Now the last step is to make use of the property$\log (\dfrac{a}{b}) = \log a - \log b$.
$ \Rightarrow \dfrac{{\log (\dfrac{{135}}{5})}}{{\log 3}}..............(8)$
Simplifying the above equation to get the final answer.
$ \Rightarrow \dfrac{{\log (27)}}{{\log 3}}..............(9)$
We can write $\log 27$as $3\log 3$.
$ \Rightarrow \dfrac{{3\log 3}}{{\log 3}}..............(9)$
Thus the final answer for this sum is $3$.
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 $\log 135$ further, though it had common factors like $5,3$. Instead we reduced $\log 405$ in such a manner that we can bring it in the form of $\log 135$ so that we could take out the common terms .This step made our sum much simpler.
Complete step-by-step answer:
First step in evaluating logarithm numerical is to simplify them. Thus the given numerical can be simplified as
$\dfrac{{\log 135 \times \log 15}}{{\log 3 \times \log 3}} - \dfrac{{\log 5 \times \log 405}}{{\log 3 \times \log 3}}........(1)$
Since the denominator of both the fractions are the same we can combine then and form a single fraction
$\dfrac{{\log 135 \times \log 15 - \log 5 \times \log 405}}{{\log 3 \times \log 3}}..............(2)$
We can now make use of the property $\log (ab) = \log a + \log b$and expand the numerator
$\dfrac{{\log 135 \times (\log 5 + \log 3) - \log 5 \times (\log 135 + \log 3)}}{{\log 3 \times \log 3}}..............(3)$
$\dfrac{{\log 135 \times \log 5 + \log 135 \times \log 3 - \log 5 \times \log 135 - \log 5 \times \log 3}}{{\log 3 \times \log 3}}..............(4)$
Removing $\log 135 \times \log 5$as they are having opposite signs.
$\dfrac{{\log 135 \times \log 3 - \log 5 \times \log 3}}{{\log 3 \times \log 3}}..............(5)$
It can be figured out from equation $3$, that $\log 3$is common in the numerator. We can take out $\log 3$as a common factor and strike off one of the $\log 3$ term from the denominator as well.
\[\dfrac{{\log 3 \times (\log 135 - \log 5)}}{{\log 3 \times \log 3}}..............(6)\]
$\dfrac{{\log 135 - \log 5}}{{\log 3}}..............(7)$
Now the last step is to make use of the property$\log (\dfrac{a}{b}) = \log a - \log b$.
$ \Rightarrow \dfrac{{\log (\dfrac{{135}}{5})}}{{\log 3}}..............(8)$
Simplifying the above equation to get the final answer.
$ \Rightarrow \dfrac{{\log (27)}}{{\log 3}}..............(9)$
We can write $\log 27$as $3\log 3$.
$ \Rightarrow \dfrac{{3\log 3}}{{\log 3}}..............(9)$
Thus the final answer for this sum is $3$.
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 $\log 135$ further, though it had common factors like $5,3$. Instead we reduced $\log 405$ in such a manner that we can bring it in the form of $\log 135$ so that we could take out the common terms .This step made our sum much simpler.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What are the major means of transport Explain each class 12 social science CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

Why should a magnesium ribbon be cleaned before burning class 12 chemistry CBSE

