Solve $\log{(x+1)}$ $+$ $\log{(x-1)}$ $\,=\,$ $\log{8}$

The addition of common logarithm of $x$ plus $1$ and common logarithm of $x$ minus $1$ equals to the common logarithm of $8$, is a given log equation in this log question and it should be solved to find the value of $x$.

Let us learn how to solve the logarithmic equation the common logarithm of $x$ plus $1$ plus common logarithm of $x$ minus $1$ is equal to the common logarithm of $8$.

Add the Common logarithmic functions

The common logarithmic functions are connected by a plus sign on the left hand side of the equation. It expresses that they should be added to find the sum of them. According to the sum rule of the logarithms, the sum of logarithms of $x$ plus $1$ and $x$ minus $1$ is equal to the logarithm of the product of them.

$\implies$ $\log{\big((x+1) \times (x-1)\big)}$ $\,=\,$ $\log{8}$

The product of two binomials inside the logarithm can be evaluated by the difference of squares formula.

$\implies$ $\log{(x^2-1^2)}$ $\,=\,$ $\log{8}$

$\implies$ $\log{(x^2-1)}$ $\,=\,$ $\log{8}$

On the left hand side of the equation, the expression inside the common logarithmic function is a quadratic expression. So, it can be understood that it is a quadratic logarithmic equation.

Eliminate the Logarithmic operation from equation

The common logarithmic operation is involved on both sides of the equation and it is possible to eliminate the common logarithmic operation from the equation by the equality property.

$\implies$ $x^2-1$ $\,=\,$ $8$

Solve the Quadratic equation by the factorisation

Now, let’s simplify the quadratic equation for solving it.

$\implies$ $x^2-1-8$ $\,=\,$ $0$

$\implies$ $x^2-9$ $\,=\,$ $0$

The first term in the expression on the left hand side of the equation is in square form. So, it is recommended to express the second term $9$ in square form.

$\implies$ $x^2-3^2$ $\,=\,$ $0$

Now, there are two terms in square form on the left hand side of the equation and they are connected by a minus sign. The difference of squares can be factored mathematically as a product of two binomials by the factorization by the difference of squares.

$\implies$ $(x+3)(x-3)$ $\,=\,$ $0$

$\implies$ $x+3\,=\, 0$ or $x-3\,=\,0$

$\,\,\,\therefore\,\,\,\,\,\,$ $x\,=\,-3$ or $x\,=\,3$

Therefore, the solution set for the common log of $x$ plus $1$ plus logarithm of $x$ minus $1$ equals to common logarithm of $8$ is $\{-3,\,3\}$.

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