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Question: How do you solve \(\ln x-\ln \left( x+2 \right)=1\)?...

How do you solve ln⁡x−ln⁡(x+2)=1\ln x-\ln \left( x+2 \right)=1?

Explanation

Solution

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 xx.

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=axy={{a}^{x}} then its equivalent logarithmic function to this is given as log⁡ay=x{{\log }_{a}}y=x. For example, we have 33=27{{3}^{3}}=27. Using the concept given above, we can rewrite this expression in the logarithmic form as log⁡327=3{{\log }_{3}}27=3.
The logarithmic quotient rule is given as
log⁡b(xy)=log⁡bx−log⁡by{{\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,
ln⁡x−ln⁡(x+2)=1 ∴ln⁡(xx+2)=1 \begin{aligned} & \ln x-\ln \left( x+2 \right)=1 \\\ & \therefore \ln \left( \dfrac{x}{x+2} \right)=1 \\\ \end{aligned}
As we have seen above, if we have the exponential function as y=axy={{a}^{x}} then its equivalent logarithmic function to this is given as log⁡ay=x{{\log }_{a}}y=x. The base of the natural logarithm is ee. So, we can convert the above equation in its exponential form as follows,
xx+2=e1\dfrac{x}{x+2}={{e}^{1}}
Now, we will simplify the above equation and solve for the variable xx in the following manner,
x=e(x+2) ∴x=ex+2e \begin{aligned} & x=e\left( x+2 \right) \\\ & \therefore x=ex+2e \\\ \end{aligned}
Shifting the terms with the variable xx to one side of the equation, we get the following,
x−ex=2ex-ex=2e
Taking xx common, we have
x(1−e)=2e ∴x=2e1−e \begin{aligned} & x\left( 1-e \right)=2e \\\ & \therefore x=\dfrac{2e}{1-e} \\\ \end{aligned}
So, the solution of the given equation is x=2e1−ex=\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.