Question

How do you solve $\log \left( x-3 \right)+\log \left( x-2 \right)=\log \left( 2x+24 \right)$ ?

Answer 1

Hint : We solve the given equation using the different identity formulas of logarithm like $\log a+\log b=\log \left( ab \right)$ , $\log a=\log b\Rightarrow a=b$ . The main step would be to eliminate the logarithm function and keep only the quadratic equation of x. we solve the equation with the help of factorisation.

Complete step-by-step answer :
We take the logarithmic identity for the given equation $\log \left( x-3 \right)+\log \left( x-2 \right)=\log \left( 2x+24 \right)$ to find the solution for x. We have $\log a+\log b=\log \left( ab \right)$ .
We operate the addition part on the left-hand side of $\log \left( x-3 \right)+\log \left( x-2 \right)=\log \left( 2x+24 \right)$ .
\log \left( x-3 \right)+\log \left( x-2 \right)=\log \left\\{ \left( x-3 \right)\left( x-2 \right) \right\\} .
The equation becomes \log \left\\{ \left( x-3 \right)\left( x-2 \right) \right\\}=\log \left( 2x+24 \right) .
Now we have to eliminate the logarithm function to find the quadratic equation of x.
We know $\log a=\log b\Rightarrow a=b$ .
Applying the rule in case of \log \left\\{ \left( x-3 \right)\left( x-2 \right) \right\\}=\log \left( 2x+24 \right) , we get
\begin{aligned} & \log \left\\{ \left( x-3 \right)\left( x-2 \right) \right\\}=\log \left( 2x+24 \right) \\\ & \Rightarrow \left( x-3 \right)\left( x-2 \right)=2x+24 \\\ \end{aligned}
Now we have a quadratic equation of x. We need to solve it.
We simplify the equation to get
$\begin{aligned} & \left( x-3 \right)\left( x-2 \right)=2x+24 \\\ & \Rightarrow {{x}^{2}}-7x-18=0 \\\ \end{aligned}$ .
We factorise the polynomial to get ${{x}^{2}}-9x+2x-18=0$ .
We take a grouping method to take commons out.
$\begin{aligned} & {{x}^{2}}-9x+2x-18=0 \\\ & \Rightarrow x\left( x-9 \right)+2\left( x-9 \right)=0 \\\ & \Rightarrow \left( x-9 \right)\left( x+2 \right)=0 \\\ \end{aligned}$
Multiplication of two numbers gives 0 which means the terms are individually 0.
The solution of the equation is $x=-2,9$ .
Now if we put $x=-2$ , the logarithmic value $\log \left( -2-3 \right)\log \left( -5 \right)$ becomes negative which is not possible.
Therefore, solution of $\log \left( x-3 \right)+\log \left( x-2 \right)=\log \left( 2x+24 \right)$ is $x=9$ .
So, the correct answer is “9”.

Note : In case of the base is not mentioned then the general solution for the base for logarithm is 10. But the base of $e$ is fixed for $\ln$ .
We also need to remember that for logarithm function there has to be a domain constraint.
For any ${{\log }_{b}}a$ , $a>0$ . This means for $\log \left( x-3 \right)+\log \left( x-2 \right)=\log \left( 2x+24 \right)$ , $\left( x-3 \right),\left( x-2 \right),\left( 2x+24 \right)>0$ . The simplified form is $x>3$ .