Question

How do you solve $\log \left( 3x+7 \right)+\log \left( x-2 \right)=1$?

Answer 1

Hint: To solve the given equation, we first need to determine the domain set, in which the solutions can exist. For this, we need to put the argument of both the logarithm terms greater than zero to obtain two inequalities. On solving the two inequalities, we will get the domain set. Then, we have to use the property $\log A+\log B=\log \left( AB \right)$ to simplify the LHS of the given equation. Then taking the antilog on both the sides, we will obtain a quadratic equation. On solving the quadratic equation, we will obtain two solutions. Finally, we have to check the solutions according to the domain set.

Complete step by step solution:
The given equation is
$\Rightarrow \log \left( 3x+7 \right)+\log \left( x-2 \right)=1$
Firstly, we determine the domain set in which the solutions of the above equation can exist. We know that the logarithm function takes only positive arguments. So from the arguments of the two logarithm terms, we obtain
$\begin{align}
  & \Rightarrow 3x+17>0 \\
 & \Rightarrow 3x>-17 \\
 & \Rightarrow x>-\dfrac{17}{3} \\
\end{align}$
Also,
$\begin{align}
  & \Rightarrow x-2>0 \\
 & \Rightarrow x>2 \\
\end{align}$
Combining the above two inequalities, we get $x>2$. Therefore, the domain set for the given equation is $x\in \left[ 2,\infty \right)$.
Now, we rewrite the given equation.
$\Rightarrow \log \left( 3x+7 \right)+\log \left( x-2 \right)=1$
We know that $\log A+\log B=\log \left( AB \right)$. Applying this property on the LHS, we get
$\Rightarrow \log \left[ \left( 3x+7 \right)\left( x-2 \right) \right]=1$
Taking anti log both the sides, we get
$\begin{align}
  & \Rightarrow \left( 3x+7 \right)\left( x-2 \right)={{10}^{1}} \\
 & \Rightarrow \left( 3x+7 \right)\left( x-2 \right)=10 \\
\end{align}$
Expanding the LHS using the distributive property, we get
\[\begin{align}
  & \Rightarrow 3x\left( x-2 \right)+7\left( x-2 \right)=10 \\
 & \Rightarrow 3{{x}^{2}}-6x+7x-14=10 \\
 & \Rightarrow 3{{x}^{2}}+x-14=10 \\
\end{align}\]
Subtracting $10$ from both sides, we get
$\begin{align}
  & \Rightarrow 3{{x}^{2}}+x-14-10=10-10 \\
 & \Rightarrow 3{{x}^{2}}+x-24=0 \\
\end{align}$
Using the middle term splitting method, we split the middle term as $x=9x-8x$ to get
\[\Rightarrow 3{{x}^{2}}+9x-8x-24=0\]
Taking $3x$ common from the first two terms, and $-8$ common from the last two terms, we get
\[\Rightarrow 3x\left( x+3 \right)-8\left( x+3 \right)=0\]
Now, taking $\left( x+3 \right)$ common, we get
$\Rightarrow \left( x+3 \right)\left( 3x-8 \right)=0$
Therefore we get $x=-3$ and $x=\dfrac{8}{3}$.
But the domain set is $x\in \left[ 2,\infty \right)$. So the solution $x=-3$ is rejected.
Hence, the final solution of the given equation is $x=\dfrac{8}{3}$.

Note: We must not forget to check the obtained solutions from the domain set. And for that we must determine the domain set in the very first step of the solution. Also, do not get confused as to why the base of logarithm is $10$. When the base of the logarithm is not specified, then it means that the base is equal to $10$.