Question

log (x - 1) + log (x + 1) = log 12​

Answer 1

$\large\text{\underline{Required value}}$

The value of $x$ that satisfies $\log(x-1)+\log(x+1)=\log12$.

$\large\text{\underline{Method}}$

The solution consists of three parts.

1. Finding the possible interval of $x$.
2. Using the properties of the logarithm.
3. Comparing the value.

$\large\text{\underline{Given equation}}$

$\implies \log(x-1)+\log(x+1)=\log12$

$\large\text{\underline{Step 1.}}$

$\large\red{\bigstar}\underline{\text{\blue{Valid domain(Power)}}}$

The first step is to find the possible interval of $x$. Sometimes logarithm equation gives solutions that are not valid.

For the logarithms to be defined, the power should remain positive.

Valid domain of $\log(x-1)$

$\implies x>1\text{ is the possible interval.}$

Valid domain of $\log(x+1)$

$\implies x>-1\text{ is the possible interval.}$

Now, their intersection $x>1$ is the possible interval.

$\large\text{\underline{Step 2.}}$

$\large\red{\bigstar}\underline{\text{\blue{Logarithm properties}}}$

The logarithm starts from the calculation of powers. Since the sum of the exponents can be interpreted as a product of two powers,

$\implies \boxed{\log_{a}x+\log_{a}y=\log_{a}xy}$

Now let's solve our problem.

$\implies \log(x-1)+\log(x+1)=\log12\ (x>1)$

$\implies\log(x-1)(x+1)=\log12\ (x>1)$

$\implies\log(x^{2}-1)=\log12\ (x>1)$

$\large\text{\underline{Step 3.}}$

$\large\red{\bigstar}\underline{\text{\blue{Increasing function}}}$

Let's consider a coordinate plane.

• $y=\log(x^{2}-1)\ (x>1)$ is always increasing.
• $y=\log12$ is a constant value.

How many solutions are there to $\log(x-1)+\log(x+1)=\log12$? Only one. The solution is unique, as the two graphs cannot intersect anymore.

This idea is used in solving the exponential and logarithmic functions, and the same goes for decreasing functions. And we can compare the values of powers.

Now let's solve our problem.

$\implies\log(x^{2}-1)=\log12\ (x>1)$

$\implies x^{2}-1=12\ (x>1)$

$\implies x^{2}=13\ (x>1)$

$\implies x=\sqrt{13}\ (\because x>1)$

And there is our answer.

$\large\text{\underline{Conclusion}}$

So, the value of $x=\sqrt{13}$.