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{align}
  & \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{align} $
Now we have a quadratic equation of x. We need to solve it.
We simplify the equation to get
 $ \begin{align}
  & \left( x-3 \right)\left( x-2 \right)=2x+24 \\
 & \Rightarrow {{x}^{2}}-7x-18=0 \\
\end{align} $ .
We factorise the polynomial to get $ {{x}^{2}}-9x+2x-18=0 $ .
We take a grouping method to take commons out.
 $ \begin{align}
  & {{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{align} $
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 $ .