Question

How do you solve $\ln \left( x \right)+\ln \left( x-2 \right)=\ln \left( 3x+14 \right)$?

Answer 1

Hint: In the given equation $\ln \left( x \right)+\ln \left( x-2 \right)=\ln \left( 3x+14 \right)$, we can clearly see that there are three logarithmic terms involved. A logarithmic function is defined only for the positive values of the argument. On applying this condition on the arguments of the three logarithmic terms to obtain the three intervals of $x$. The intersection of those three intervals will be the domain of the solution of the given equation. Then we have to use the logarithm properties $\ln A+\ln B=\ln AB$ and $\ln A-\ln B=\ln \dfrac{A}{B}$ to simplify the equation and then taking the antilog on both the sides of the equation obtained, we will get a quadratic equation. On solving the quadratic equation we will get two solutions. Finally, we have to reject the solutions according to the domain of the solutions and the leftover solution will be the final answer.

Complete step-by-step answer:
The equation to be solved is given in the question as
$\ln \left( x \right)+\ln \left( x-2 \right)=\ln \left( 3x+14 \right)..........(i)$
Before solving the equation, we determine the domain of the solutions of the above equation. We know that the logarithm function is defined only for the positive values of domain. From the three logarithm terms present in the above equation, we obtain the following conditions
For $\ln \left( x \right):$
$\Rightarrow x>0$
For $\ln \left( x-2 \right):$
$\begin{align}
  & \Rightarrow x-2>0 \\
 & \Rightarrow x>2 \\
\end{align}$
For $\ln \left( 3x+14 \right):$
$\begin{align}
  & \Rightarrow 3x+14>0 \\
 & \Rightarrow 3x>-14 \\
 & \Rightarrow x>-\dfrac{14}{3} \\
\end{align}$
Taking the intersection of the above three inequalities, we obtain the domain of the solutions as
$\begin{align}
  & \Rightarrow x>2 \\
 & \Rightarrow x\in \left( 2,\infty \right) \\
\end{align}$
Now, we again consider the given equation
$\Rightarrow \ln \left( x \right)+\ln \left( x-2 \right)=\ln \left( 3x+14 \right)$
We know the property $\ln A+\ln B=\ln AB$. So the above equation can be written as
\[\Rightarrow \ln \left( x\left( x-2 \right) \right)=\ln \left( 3x+14 \right)\]
Subtracting $\ln \left( 3x+14 \right)$ from both sides
\[\Rightarrow \ln \left( x\left( x-2 \right) \right)-\ln \left( 3x+14 \right)=0\]
From the property $\ln A-\ln B=\ln \dfrac{A}{B}$ the above equation can be written as
\[\Rightarrow \ln \left( \dfrac{x\left( x-2 \right)}{\left( 3x+14 \right)} \right)=0\]
Now, the base of the logarithmic function here is $e$ which is not equal to one. Also, since according to the domain of solutions, we have $x>2$ which makes the argument \[\dfrac{x\left( x-2 \right)}{\left( 3x+14 \right)}>0\]. So we can take the antilog on both sides of the above equation to get
\[\begin{align}
  & \Rightarrow \dfrac{x\left( x-2 \right)}{\left( 3x+14 \right)}={{e}^{0}} \\
 & \Rightarrow \dfrac{x\left( x-2 \right)}{\left( 3x+14 \right)}=1 \\
 & \Rightarrow x\left( x-2 \right)=3x+14 \\
 & \Rightarrow {{x}^{2}}-2x=3x+14 \\
 & \Rightarrow {{x}^{2}}-5x-14=0 \\
\end{align}\]
Splitting the middle term as $-5x=2x-7x$, we get
$\begin{align}
  & \Rightarrow {{x}^{2}}+2x-7x+14=0 \\
 & \Rightarrow x\left( x+2 \right)-7\left( x+2 \right)=0 \\
 & \Rightarrow \left( x+2 \right)\left( x-7 \right)=0 \\
 & \Rightarrow x=-2,x=7 \\
\end{align}$
Now, from the domain of the solution, we have $x\in \left( 2,\infty \right)$. But the solution $x=-2$ does not belong to the domain. So we reject the solution $x=-2$ and hence the final solution of the given equation is $x=7$.

Note: Do not forget to determine the domain of the solutions of the equation given. The domain of solutions is determined from the form of the equation given in the question as it is. Do not make any manipulation in the given equation before determining the domain. Also, do not forget to reject the solutions on the basis of the domain.