Question

How do you solve $ {{\log }_{2}}\left( x+7 \right)+{{\log }_{2}}\left( x+8 \right)=1 $ ?

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 }_{m}}a=y\Rightarrow a={{m}^{y}} $ . 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 }_{2}}\left( x+7 \right)+{{\log }_{2}}\left( x+8 \right)=1 $ to find the solution for x. We have $ \log a+\log b=\log \left( ab \right) $ .
We operate the addition part in the left-hand side of $ {{\log }_{2}}\left( x+7 \right)+{{\log }_{2}}\left( x+8 \right)=1 $ .
 $ {{\log }_{2}}\left( x+7 \right)+{{\log }_{2}}\left( x+8 \right)={{\log }_{2}}\left\{ \left( x+7 \right)\left( x+8 \right) \right\} $ .
The equation becomes $ {{\log }_{2}}\left\{ \left( x+7 \right)\left( x+8 \right) \right\}=1 $ .
Now we have to eliminate the logarithm function to find the quadratic equation of x.
We know $ {{\log }_{m}}a=y\Rightarrow a={{m}^{y}} $ .
Applying the rule in case of $ {{\log }_{2}}\left\{ \left( x+7 \right)\left( x+8 \right) \right\}=1 $ , we get
 $
   {{\log }_{2}}\left\{ \left( x+7 \right)\left( x+8 \right) \right\}=1 \\
  \Rightarrow \left( x+7 \right)\left( x+8 \right)={{2}^{1}}=2 \;
 $
Now we have a quadratic equation of x. We need to solve it.
We simplify the equation to get $ {{x}^{2}}+15x+54=0 $ .
We factorise the polynomial to get $ {{x}^{2}}+9x+6x+54=0 $ .
We take a grouping method to take commons out.
 $
   {{x}^{2}}+9x+6x+54=0 \\
  \Rightarrow x\left( x+9 \right)+6\left( x+9 \right)=0 \\
  \Rightarrow \left( x+9 \right)\left( x+6 \right)=0 \;
 $
Multiplication of two numbers gives 0 which means the terms are individually 0.
The solution of the equation is $ x=-6,-9 $ .
Now if we put $ x=-9 $ , the logarithmic values become negative which is not possible.
Therefore, solution of $ {{\log }_{2}}\left( x+7 \right)+{{\log }_{2}}\left( x+8 \right)=1 $ is $ x=-6 $
So, the correct answer is “ $ x=-6 $ ”.

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 }_{2}}\left( x+7 \right),{{\log }_{2}}\left( x+8 \right) $ , $ \left( x+7 \right)>0 $ and $ \left( x+8 \right)>0 $ .
The simplified form is $ x>-8 $ .