
How do you solve $ \ln \left( x+1 \right)-\ln \left( x \right)=2 $ ?
Answer
564.3k+ views
Hint:
We will use the identity of logarithm which is $ \ln a-\ln b=\ln \left( \dfrac{a}{b} \right) $ . We will see the relation between the natural logarithm function and the natural exponential function. Then we will convert the logarithm function into the exponential function. After that, we will solve the equation to obtain the value of $ x $ in terms of $ e $.
Complete step by step answer:
The given equation is the following,
$ \ln \left( x+1 \right)-\ln \left( x \right)=2 $
We have a property of the logarithm function which tells us that
$ \ln a-\ln b=\ln \left( \dfrac{a}{b} \right) $
Therefore, we can rewrite the given equation using the above mentioned property in the following manner,
$ \ln \left( \dfrac{x+1}{x} \right)=2 $
We know that the logarithm function is the inverse of the exponential function. Rather, we have the following definition,
If $ \ln a=b $ , then $ a={{e}^{b}} $ .
Using this definition, we can convert the given logarithm function into exponential function in the following manner,
$ \begin{align}
& \ln \left( \dfrac{x+1}{x} \right)=2 \\
& \therefore \dfrac{x+1}{x}={{e}^{2}} \\
\end{align} $
Now, we will rearrange the above equation to solve for $ x $ as follows,
$ \begin{align}
& x+1=x{{e}^{2}} \\
& \therefore x{{e}^{2}}-x=1 \\
\end{align} $
Taking $ x $ common, we get the following equation,
$ \begin{align}
& x\left( {{e}^{2}}-1 \right)=1 \\
& \therefore x=\dfrac{1}{{{e}^{2}}-1} \\
\end{align} $
This is the solution for the given equation.
Note:
The number $ e $ is Euler's number. The approximate value of this number is 2.71828. We can further substitute the value of $ e $ in the solution. So we get, $ x=\dfrac{1}{{{\left( 2.71828 \right)}^{2}}-1} $ . Upon solving this equation, we obtain a value of $ x=0.1565 $ . It is important to know the properties and identities of the logarithm function and the exponential function. Doing the calculations explicitly allows us to keep a track of the properties we are using for simplifying the equation. It also minimizes any chances of making errors.
We will use the identity of logarithm which is $ \ln a-\ln b=\ln \left( \dfrac{a}{b} \right) $ . We will see the relation between the natural logarithm function and the natural exponential function. Then we will convert the logarithm function into the exponential function. After that, we will solve the equation to obtain the value of $ x $ in terms of $ e $.
Complete step by step answer:
The given equation is the following,
$ \ln \left( x+1 \right)-\ln \left( x \right)=2 $
We have a property of the logarithm function which tells us that
$ \ln a-\ln b=\ln \left( \dfrac{a}{b} \right) $
Therefore, we can rewrite the given equation using the above mentioned property in the following manner,
$ \ln \left( \dfrac{x+1}{x} \right)=2 $
We know that the logarithm function is the inverse of the exponential function. Rather, we have the following definition,
If $ \ln a=b $ , then $ a={{e}^{b}} $ .
Using this definition, we can convert the given logarithm function into exponential function in the following manner,
$ \begin{align}
& \ln \left( \dfrac{x+1}{x} \right)=2 \\
& \therefore \dfrac{x+1}{x}={{e}^{2}} \\
\end{align} $
Now, we will rearrange the above equation to solve for $ x $ as follows,
$ \begin{align}
& x+1=x{{e}^{2}} \\
& \therefore x{{e}^{2}}-x=1 \\
\end{align} $
Taking $ x $ common, we get the following equation,
$ \begin{align}
& x\left( {{e}^{2}}-1 \right)=1 \\
& \therefore x=\dfrac{1}{{{e}^{2}}-1} \\
\end{align} $
This is the solution for the given equation.
Note:
The number $ e $ is Euler's number. The approximate value of this number is 2.71828. We can further substitute the value of $ e $ in the solution. So we get, $ x=\dfrac{1}{{{\left( 2.71828 \right)}^{2}}-1} $ . Upon solving this equation, we obtain a value of $ x=0.1565 $ . It is important to know the properties and identities of the logarithm function and the exponential function. Doing the calculations explicitly allows us to keep a track of the properties we are using for simplifying the equation. It also minimizes any chances of making errors.
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