
How do you evaluate ${{\log }_{3}}7$ using the change of base formula?
Answer
538.2k+ views
Hint: The logarithm is used to convert a large or very small number into the understandable domain. For the theorem to work the usual conditions of logarithm will have to follow. We also need to remember that for logarithm function there has to be a domain constraint. The range in the usual case is the whole real line.
Complete step-by-step solution:
Let an arbitrary logarithmic function be $A={{\log }_{b}}a$. The conditions for the expression to be logical is $a,b>0;b\ne 1$.
We have $\log {{x}^{a}}=a\log x$. The power value of $a$ goes as a multiplication with $\log x$.
We also have the identity for logarithm where ${{\log }_{y}}x=\dfrac{{{\log }_{m}}x}{{{\log }_{m}}y}$.
The new base can be anything but the condition is that the base for both denominator and the numerator have to be the same.
Using the formula, we can break the logarithm ${{\log }_{3}}7$ taking the base as exponent $e$.
We have $\ln a={{\log }_{e}}a$.
Therefore, ${{\log }_{3}}7=\dfrac{{{\log }_{e}}7}{{{\log }_{e}}3}=\dfrac{\ln 7}{\ln 3}$.
Now we can put these values from the calculator to find the value for ${{\log }_{3}}7$.
We have $\ln 3=1.1,\ln 7=1.946$. This gives ${{\log }_{3}}7=\dfrac{\ln 7}{\ln 3}=\dfrac{1.946}{1.1}=1.098$.
The value of ${{\log }_{3}}7$ is $1.098$.
Note: There are some particular rules that we follow in case of finding the condensed form of logarithm. We first apply the power property first. Then we identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power. Then we apply the product property. Rewrite sums of logarithms as the logarithm of a product. We also have the quotient property rules.
Complete step-by-step solution:
Let an arbitrary logarithmic function be $A={{\log }_{b}}a$. The conditions for the expression to be logical is $a,b>0;b\ne 1$.
We have $\log {{x}^{a}}=a\log x$. The power value of $a$ goes as a multiplication with $\log x$.
We also have the identity for logarithm where ${{\log }_{y}}x=\dfrac{{{\log }_{m}}x}{{{\log }_{m}}y}$.
The new base can be anything but the condition is that the base for both denominator and the numerator have to be the same.
Using the formula, we can break the logarithm ${{\log }_{3}}7$ taking the base as exponent $e$.
We have $\ln a={{\log }_{e}}a$.
Therefore, ${{\log }_{3}}7=\dfrac{{{\log }_{e}}7}{{{\log }_{e}}3}=\dfrac{\ln 7}{\ln 3}$.
Now we can put these values from the calculator to find the value for ${{\log }_{3}}7$.
We have $\ln 3=1.1,\ln 7=1.946$. This gives ${{\log }_{3}}7=\dfrac{\ln 7}{\ln 3}=\dfrac{1.946}{1.1}=1.098$.
The value of ${{\log }_{3}}7$ is $1.098$.
Note: There are some particular rules that we follow in case of finding the condensed form of logarithm. We first apply the power property first. Then we identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power. Then we apply the product property. Rewrite sums of logarithms as the logarithm of a product. We also have the quotient property rules.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

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

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

