Solving the equation for $x$ given that $x \in \mathbb{R}, x > 0$

Question

Given that the product of $\log(x+3)$ and $\log(x-3)$ is equal to $3$; the logarithms are to the base of 3.

$$ \log_3(x+3)\log_3(x-3) = 3 $$

Someone to help me solve for $x$?

Answer 1

The equation $$f(x)=\log_3(x+3) \times \log_3(x-3) - 3=0$$ cannot be solved analytically. Only numerical methods could be used for solving it. Plotting the function shows that there is a root in the range $7\lt x \lt 8$. A good candidate is Newton iterative scheme $$x_{n+1}=x_n-\frac {f(x_n)}{f'(x_n)}$$ Applied to this function and starting iterations at $x_0=7$ gives the following iterates : $7.60031$, $7.62747$, $7.62751$ which is the solution for six significant figures.

Added later to this answer

Even without plotting the function, it is obvious that the solution of the equation is larger than $6$ since $f(6)=-1$. So, iterations can start at this value and the successive Newton iterates are $7.41250$, $7.62449$, $7.62751$