Question

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

Answer 1

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.