How to solve $\log_2(x) +3 = \log_3(x+2)$

Question

Hi Math Stack Exchange Communities, I am new here. I have a question regarding logarithm solving.

Let's say I have this equation:

$$\log_2 (x) +3 = \log_3 (x+2)$$

How can I solve this kind of equation?

Thank you!

Answer 1

Changing to natural logarithms, the function you look the zero for write$$f(x)=\frac{\log (x)}{\log (2)}-\frac{\log (x+2)}{\log (3)}+3$$ Its derivative $$f'(x)=\frac{1}{x \log (2)}-\frac{1}{(x+2) \log (3)}$$ would cancel for $x=\frac{2 \log (3)}{\log (2)-\log (3)}<0$ and you could easily show that $f'(x) >0$ for all $x>0$.

If $x\to 0$, $f(x)\to -\infty$ and by inspection $f(1)=2$; so, there is a root between $0$ and $1$.

At this point, we need some numerical method and Newton would probably be the simplest to use. The problem is that, starting with $x_0=1$ will give a first iterate which will be negative (outside the acceptable range). Plotting the function would be the simplest way the locate more or less accurately the root. Otherwise, let us tabulate some values $$\left( \begin{array}{cc} x & f(x) \\ 0.1 & -0.997269 \\ 0.2 & -0.039613 \\ 0.3 & +0.504888 \\ 0.4 & +0.881186 \end{array}\right)$$

So, let us use $x_0=0.2$. Newton iterates will then be $$\left( \begin{array}{cc} 1 & 0.2058256595809923861 \\ 2 & 0.2059161927467302404 \\ 3 & 0.2059162137902695078 \\ 4 & 0.2059162137902706441 \end{array} \right)$$