Question

How do you solve \[\log (x-1)+\log (x+1)=2\log (x+2)\]?

Answer 1

Hint: To solve the given problems we should know some of the properties of logarithm, they are given below. The first property we should know is the addition of logarithm, \[\log a+\log b=\log ab\]. The second property we should know is \[\ln {{a}^{b}}=b\ln a\]. We should also know that the argument of logarithm should always be positive, that is in \[\log a\] the range of a must be \[\left( 0,\infty \right)\]. otherwise, the logarithm becomes undefined.

Complete step-by-step solution:
We are given the equation \[\log (x-1)+\log (x+1)=2\log (x+2)\]. Using the property \[\log a+\log b=\log ab\] on the LHS of the above equation, we get
\[\Rightarrow \log \left( (x-1)(x+1) \right)=2\log (x+2)\]
Using the property \[\ln {{a}^{b}}=b\ln a\] on the RHS of the above equation, it can be expressed as
\[\Rightarrow \log \left( (x-1)(x+1) \right)=\log {{(x+2)}^{2}}\]
If the log of two values is equal, then the values must be the same. Hence,
\[\left( (x-1)(x+1) \right)={{(x+2)}^{2}}\]
Expanding the brackets on both sides of the above equation, we get
\[\Rightarrow {{x}^{2}}-1={{x}^{2}}+4x+4\]
Simplifying the above equation, it can be expressed as
\[\Rightarrow -1=4x+4\]
Solving the above equation, we get \[x=\dfrac{-5}{4}\].
But if substitute \[x=\dfrac{-5}{4}\] in the given equation, \[\log (x-1)\And \log (x+1)\] becomes undefined. Hence \[x=\dfrac{-5}{4}\] is not the solution of the equation. So, the equation has no solution.

Note: To solve these types of problems, one should know the properties of the logarithm. The properties we used here are \[\ln {{a}^{b}}=b\ln a\], and \[\log a+\log b=\log ab\]. As these are properties, they can be used in both directions. After solving the logarithmic problems, we must verify whether the solution is satisfying the conditions for the base/ argument of logarithm or not. If it's not then we have to exclude it. As here \[x=\dfrac{-5}{4}\] was making \[\log (x-1)\And \log (x+1)\] undefined, so we had to exclude it.