How do you solve $\ln x-\ln \left( x+2 \right)=1$?
Answer
Verified
439.5k+ views
Hint: We will look at the definition of logarithmic function. Then we will look at the logarithmic quotient rule. We will use this rule to simplify the given expression. Then we will convert the logarithmic function into its exponential form. After that we will simplify the equation obtained and solve it for the variable $x$.
Complete step by step answer:
The logarithmic function is defined as the inverse of the exponential function. So, if we have the exponential function as $y={{a}^{x}}$ then its equivalent logarithmic function to this is given as ${{\log }_{a}}y=x$. For example, we have ${{3}^{3}}=27$. Using the concept given above, we can rewrite this expression in the logarithmic form as ${{\log }_{3}}27=3$.
The logarithmic quotient rule is given as
${{\log }_{b}}\left( \dfrac{x}{y} \right)={{\log }_{b}}x-{{\log }_{b}}y$
Using this rule, we can write the given expression as the following,
$\begin{align}
& \ln x-\ln \left( x+2 \right)=1 \\
& \therefore \ln \left( \dfrac{x}{x+2} \right)=1 \\
\end{align}$
As we have seen above, if we have the exponential function as $y={{a}^{x}}$ then its equivalent logarithmic function to this is given as ${{\log }_{a}}y=x$. The base of the natural logarithm is $e$. So, we can convert the above equation in its exponential form as follows,
$\dfrac{x}{x+2}={{e}^{1}}$
Now, we will simplify the above equation and solve for the variable $x$ in the following manner,
$\begin{align}
& x=e\left( x+2 \right) \\
& \therefore x=ex+2e \\
\end{align}$
Shifting the terms with the variable $x$ to one side of the equation, we get the following,
$x-ex=2e$
Taking $x$ common, we have
$\begin{align}
& x\left( 1-e \right)=2e \\
& \therefore x=\dfrac{2e}{1-e} \\
\end{align}$
So, the solution of the given equation is $x=\dfrac{2e}{1-e}$.
Note:
We should be familiar with the logarithmic functions and exponential functions. The relation between these two functions is very useful in simplification and calculation. Like the logarithmic quotient rule, there are other rules like product rule, power rule, base switch rule etc. These rules are important for simplifying equations.
Complete step by step answer:
The logarithmic function is defined as the inverse of the exponential function. So, if we have the exponential function as $y={{a}^{x}}$ then its equivalent logarithmic function to this is given as ${{\log }_{a}}y=x$. For example, we have ${{3}^{3}}=27$. Using the concept given above, we can rewrite this expression in the logarithmic form as ${{\log }_{3}}27=3$.
The logarithmic quotient rule is given as
${{\log }_{b}}\left( \dfrac{x}{y} \right)={{\log }_{b}}x-{{\log }_{b}}y$
Using this rule, we can write the given expression as the following,
$\begin{align}
& \ln x-\ln \left( x+2 \right)=1 \\
& \therefore \ln \left( \dfrac{x}{x+2} \right)=1 \\
\end{align}$
As we have seen above, if we have the exponential function as $y={{a}^{x}}$ then its equivalent logarithmic function to this is given as ${{\log }_{a}}y=x$. The base of the natural logarithm is $e$. So, we can convert the above equation in its exponential form as follows,
$\dfrac{x}{x+2}={{e}^{1}}$
Now, we will simplify the above equation and solve for the variable $x$ in the following manner,
$\begin{align}
& x=e\left( x+2 \right) \\
& \therefore x=ex+2e \\
\end{align}$
Shifting the terms with the variable $x$ to one side of the equation, we get the following,
$x-ex=2e$
Taking $x$ common, we have
$\begin{align}
& x\left( 1-e \right)=2e \\
& \therefore x=\dfrac{2e}{1-e} \\
\end{align}$
So, the solution of the given equation is $x=\dfrac{2e}{1-e}$.
Note:
We should be familiar with the logarithmic functions and exponential functions. The relation between these two functions is very useful in simplification and calculation. Like the logarithmic quotient rule, there are other rules like product rule, power rule, base switch rule etc. These rules are important for simplifying equations.
Recently Updated Pages
Class 11 Question and Answer - Your Ultimate Solutions Guide
Master Class 11 English: Engaging Questions & Answers for Success
Master Class 11 Computer Science: Engaging Questions & Answers for Success
Master Class 11 Maths: Engaging Questions & Answers for Success
Master Class 11 Social Science: Engaging Questions & Answers for Success
Master Class 11 Economics: Engaging Questions & Answers for Success
Trending doubts
10 examples of friction in our daily life
What problem did Carter face when he reached the mummy class 11 english CBSE
Difference Between Prokaryotic Cells and Eukaryotic Cells
State and prove Bernoullis theorem class 11 physics CBSE
What organs are located on the left side of your body class 11 biology CBSE
Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE