Question

How do you solve \[\log x+\log (x+21)=2\]?

Answer 1

Hint: To solve the given equation we will need to use some properties and formulas, first are the properties of logarithm which states that \[\log a+\log b=\log \left( ab \right)\], here \[a > 0\And b > 0\], and other is \[{{\log }_{a}}m=n\Rightarrow {{a}^{n}}=m\]. The base of\[\log \]is 10. Also, we should know that for the quadratic equation \[a{{x}^{2}}+bx+c=0\], here \[a,b,c\in R\] using the formula method the roots of the equation are \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].

Complete step by step answer:
The given equation is \[\log x+\log (x+21)=2\].
We will use a property of logarithm which states that \[\log a+\log b=\log \left( ab \right)\], using this in the above equation we get,
\[\Rightarrow \log \left( x(x+21) \right)=2\]
As the base of \[\log \] is 10, we use another property which states that \[{{\log }_{a}}m=n\Rightarrow {{a}^{n}}=m\], so we get,
\[\Rightarrow {{x}^{2}}+21x=100\]
subtracting 100 from both sides of the equation we get,
\[\Rightarrow {{x}^{2}}+21x-100=0\]
For the quadratic equation, \[a{{x}^{2}}+bx+c=0\], using the formula method the roots of the equation are \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\], comparing the equation \[{{x}^{2}}+21x-100=0\] with \[a{{x}^{2}}+bx+c=0\], we get \[a=1,b=21,c=-100\].
Using these values in the above formula, we get the roots of the equation as,
\[\begin{align}
  & \Rightarrow x=\dfrac{-21\pm \sqrt{841}}{2} \\
 & \Rightarrow x=\dfrac{-21\pm 29}{2} \\
\end{align}\]
 \[\Rightarrow x=\dfrac{8}{2}\] or \[x=\dfrac{-50}{2}\]
\[\therefore x=4\] or \[x=-25\]

But, if we consider the values \[x=-25\] then the given equation becomes, \[\log (-25)+\log (-25+21)=2\] this is not acceptable as the argument of logarithm must be always positive. So, \[x=-25\] is rejected and the only solution is \[x=4\].

Note: For problems involving logarithm, it is very important to check whether the solution is satisfying the condition for logarithm or not? If any solution is not satisfying, we have to reject it. We can check whether the solution is correct or not, as we have rejected \[x=-25\], we have to check for only \[x=4\]. Substitute \[x=4\] in the Left-Hand Side of the equation we get,
\[\begin{align}
  & LHS=\log 4+\log (4+21) \\
 & \Rightarrow \log 4+\log 25 \\
\end{align}\]
Using logarithm property \[\log a+\log b=\log \left( ab \right)\], we get,
\[\begin{align}
  & \Rightarrow \log (4\times 25) \\
 & \Rightarrow \log (100)=\log ({{10}^{2}})=2\log 10=2 \\
 & \Rightarrow RHS \\
\end{align}\]
As LHS = RHS, our solution is correct.