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

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

Explanation

Solution

Hint : We solve the given equation using the different identity formulas of logarithm like log⁡a+log⁡b=log⁡(ab)\log a+\log b=\log \left( ab \right) , log⁡ma=y⇒a=my{{\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(x+7)+log⁡2(x+8)=1{{\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⁡(ab)\log a+\log b=\log \left( ab \right) .
We operate the addition part in the left-hand side of log⁡2(x+7)+log⁡2(x+8)=1{{\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⁡ma=y⇒a=my{{\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 x2+15x+54=0{{x}^{2}}+15x+54=0 .
We factorise the polynomial to get x2+9x+6x+54=0{{x}^{2}}+9x+6x+54=0 .
We take a grouping method to take commons out.
x2+9x+6x+54=0 ⇒x(x+9)+6(x+9)=0 ⇒(x+9)(x+6)=0   {{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,−9x=-6,-9 .
Now if we put x=−9x=-9 , the logarithmic values become negative which is not possible.
Therefore, solution of log⁡2(x+7)+log⁡2(x+8)=1{{\log }_{2}}\left( x+7 \right)+{{\log }_{2}}\left( x+8 \right)=1 is x=−6x=-6
So, the correct answer is “ x=−6x=-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 ee is fixed for ln⁡\ln .
We also need to remember that for logarithm function there has to be a domain constraint.
For any log⁡ba{{\log }_{b}}a , a>0a>0 . This means for log⁡2(x+7),log⁡2(x+8){{\log }_{2}}\left( x+7 \right),{{\log }_{2}}\left( x+8 \right) , (x+7)>0\left( x+7 \right)>0 and (x+8)>0\left( x+8 \right)>0 .
The simplified form is x>−8x>-8 .