Nernst Equation - "Natural Log" or "Log Base 10"

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Nernst_Equation#mjx-eqn-6

https://www.quora.com/How-does-converting-ln-to-log-in-the-Nernst-equation-involve-a-value-of-2-303-in-it?share=1

E q u a l i b r i u m o f I o n " X " = E_{x} = \frac{\frac{8.314 J o u l e s}{1 K e l v i n \cdot m o l} * (37 ° C + 273.15) ° K}{C h a r g e o f I o n (x) * \frac{96485 J o u l e s}{V o l t s \cdot m o l}} * l n (\frac{[X]_{o u t s i d e} (M o l a r)}{[X]_{i n s i d e} (M o l a r)})

Somehow it's something to do with simplifying part of the Nernst Equation to a constant. They "pre-compute" the gas constant to faraday constant ratio , and assume 37°C, physiological temperature.

They simplify this part of the Nernst Equation:

\frac{\frac{8.314 J o u l e s}{1 K e l v i n \cdot m o l} * ° K}{C h a r g e o f I o n (z) * \frac{96485 J o u l e s}{V o l t s \cdot m o l}}

1.) Separate out the Charge of the Ion Variable:

(\frac{1}{C h a r g e o f I o n (z)}) * (\frac{\frac{8.314 J o u l e s}{1 K e l v i n \cdot m o l} * (37 ° C + 273.15) ° K}{\frac{96485 J o u l e s}{V o l t s \cdot m o l}})

2.) Change Division to Multiplication of Inverse:

(\frac{1}{C h a r g e o f I o n (z)}) * (\frac{8.314 J o u l e s}{1 K e l v i n \cdot m o l} * (310.15 ° K)) * (\frac{V o l t s \cdot m o l}{96485 J o u l e s})

3.) Cancel Units:

(\frac{1}{C h a r g e o f I o n (z)}) * (\frac{8.314 J o u l e s}{1 K e l v i n \cdot m o l} * (310.15 ° K)) * (\frac{V o l t s \cdot m o l}{96485 J o u l e s})

4.) Simplify:

(\frac{1}{C h a r g e o f I o n (z)}) * (8.314 * 310.15) * (\frac{V o l t s}{96485})

5.) We are left with:

\frac{0.02672526403067834 V o l t s}{C h a r g e o f I o n (z)} * \frac{1000 m V}{1 V o l t} = \frac{26.72526403067834 m V}{C h a r g e o f I o n (z)}

6.) Rewriting the Nernst Equation with our Simplified Constants:

E q u a l i b r i u m o f I o n " X " = E_{x} = \frac{26.72526403067834 m V}{C h a r g e o f I o n (x)} * l n (\frac{[X]_{o u t s i d e} (M o l a r)}{[X]_{i n s i d e} (M o l a r)})

7.) Here is where I get confused though. They are changing the base of the logarithm from "e" to 10. But I don't understand "why" they are doing it. This change of base stuff still confuses me:

T h e N o r m a l L o g B a s e C h a n g e P r o p e r t y = l o g_{b} (x) = \frac{l o g_{c} (x)}{l o g_{c} (b)}

Our base is "e" because we started with the natural log version

\begin{matrix} L e t b = e \\ L e t c = 10 \end{matrix}

l o g_{e} (x) = \frac{l o g_{10} (x)}{l o g_{10} (e)}

Since these are all equivalent:

l o g_{e} (x) = l n (x) = (l o g_{e} 10 * l o g_{10} (x))

∴ l o g_{e} (10) = l n (10) = 2.302585092994046

We can rewrite it as:

l n (x) = (2.302585092994046) * l o g_{10} (x)

8.) Finally, we can rewrite the Nernst Equation in the "log base 10" form:

E_{x} = \frac{26.72526403067834 m V}{C h a r g e o f I o n (x)} * (2.302585092994046) * l o g_{10} (\frac{[X]_{o u t s i d e} (M o l a r)}{[X]_{i n s i d e} (M o l a r)})

This is the only rational I could find on duckduckgo images. Its just a preference/convenience thing?