
How do you solve \[\ln x - \ln 2 = 0\]?
Answer
538.8k+ views
Hint:To solve the expression for logarithmic terms, we should know the logarithmic properties like multiplication division and when we remove the log term then the equation R.H.S will get the term of “e” and the whole expression will goes to the power the this term “e”, value of this function is 2.71828, if needed then we can solve it by using this value.
Formulae Used:
\[ \Rightarrow \ln a - \ln b = \ln \dfrac{a}{b}\]
\[
\Rightarrow \ln x = y \\
then \\
\Rightarrow x = {e^y} \\
\]
Complete step by step solution:
The given question needs to solve the expression \[\ln x - \ln 2 = 0\]
Using the property of subtraction in logarithm we can say:
\[ \Rightarrow \ln a - \ln b = \ln \dfrac{a}{b}\]
Using this property in our question we can write the given expression as:
\[
\Rightarrow \ln x - \ln 2 = 0 \\
\Rightarrow \ln \dfrac{x}{2} = 0 \\
\]
Now using the second property which is when you remove the logarithm function then the expression on the second hand of equation will goes to the power of “e”, this property can be written as:
\[
\Rightarrow \ln x = y \\
then \\
\Rightarrow x = {e^y} \\
\]
Using this property in our question we can solve as:
\[
\Rightarrow \ln \dfrac{x}{2} = 0 \\
\Rightarrow \dfrac{x}{2} = {e^0} \\
\Rightarrow \dfrac{x}{2} = 1 \\
\Rightarrow x = 1 \times 2 = 2 \\
\]
Hence we obtained the final answer for the given expression.
Additional Information: The given expression needs to be solved as steps are used above, these are standard steps and need to be used for getting the final answer, the final answer can be cross checked by putting the value of the variable in the equation given in question.
Note: Here the given expression can also be solved by differentiating the equation, on differentiating the log term will be simplified to the normal integer and variable, here using differentiation the final answer would not be changed, you will obtain the same result as we get here.
Formulae Used:
\[ \Rightarrow \ln a - \ln b = \ln \dfrac{a}{b}\]
\[
\Rightarrow \ln x = y \\
then \\
\Rightarrow x = {e^y} \\
\]
Complete step by step solution:
The given question needs to solve the expression \[\ln x - \ln 2 = 0\]
Using the property of subtraction in logarithm we can say:
\[ \Rightarrow \ln a - \ln b = \ln \dfrac{a}{b}\]
Using this property in our question we can write the given expression as:
\[
\Rightarrow \ln x - \ln 2 = 0 \\
\Rightarrow \ln \dfrac{x}{2} = 0 \\
\]
Now using the second property which is when you remove the logarithm function then the expression on the second hand of equation will goes to the power of “e”, this property can be written as:
\[
\Rightarrow \ln x = y \\
then \\
\Rightarrow x = {e^y} \\
\]
Using this property in our question we can solve as:
\[
\Rightarrow \ln \dfrac{x}{2} = 0 \\
\Rightarrow \dfrac{x}{2} = {e^0} \\
\Rightarrow \dfrac{x}{2} = 1 \\
\Rightarrow x = 1 \times 2 = 2 \\
\]
Hence we obtained the final answer for the given expression.
Additional Information: The given expression needs to be solved as steps are used above, these are standard steps and need to be used for getting the final answer, the final answer can be cross checked by putting the value of the variable in the equation given in question.
Note: Here the given expression can also be solved by differentiating the equation, on differentiating the log term will be simplified to the normal integer and variable, here using differentiation the final answer would not be changed, you will obtain the same result as we get here.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

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

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

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

