I have this equation that where im trying to convert the ln base to log10

$$y = 3101.420903 \ln(x) - 8588.741253$$

Id be really grateful for some pointers on how to do this.


  • 1
    $\begingroup$ Google "change of base formula". $\endgroup$ Mar 14, 2014 at 10:17
  • $\begingroup$ yes, Ive done that, but im struggling to apply the examples I have found to my formula. $\endgroup$
    – Ke.
    Mar 14, 2014 at 10:24

3 Answers 3


There is a very general formula for changing the base of logarithms:

$$\log_a(x) = \frac{\log_b(x)}{\log_b(a)}.$$

In your case, $a=10$ and $b=e$.

  • $\begingroup$ Sorry, im a bit new to these symbols. Is there anywhere I can go for an explanation of these? $\endgroup$
    – Ke.
    Mar 14, 2014 at 10:23
  • $\begingroup$ Which symbol here is new? $\endgroup$
    – 5xum
    Mar 14, 2014 at 10:28
  • $\begingroup$ Im just not familiar with the notation youre using. What are the dollar signs for for instance? Why is there an underscore connected to log? What do these mean? How do I apply these to my formula? $\endgroup$
    – Ke.
    Mar 14, 2014 at 10:31
  • $\begingroup$ The dollar signs should not be visible here. It seems MathJax is not working form you. This is how it should look like: i.stack.imgur.com/h4VmF.png $\endgroup$
    – 5xum
    Mar 14, 2014 at 10:33
  • $\begingroup$ What is this mathjax, do I have to install it? More confused :/ $\endgroup$
    – Ke.
    Mar 14, 2014 at 10:35

I can never remember the "change-of-base" formula for logs, so I routinely just work it out, like this.

Start with the following identity (which follows directly from the definition of $\log_a$):

$$x = a^{\log_a (x)}.$$

Now, take $\log_b$ of both sides:

$$\log_b (x) = \log_b (a^{\log_a (x)}).$$

Finally, use the general identity $\log_r (s^t) = t \log_r(s)$ to conclude from the equation above that

$$\log_b(x) = \log_a(x) \log_b (a).$$

This is your "change-of-base" formula.

Flipping the terms in the RHS product results in a form of this formula that may be easier to remember (though, as I said, I never manage to do this):

$$\log_b(x) = \log_b (a) \log_a(x).$$

Note how now the RHS appears as though it were obtained from the LHS by "inserting" the (entirely meaningless!) expression $(a) \log_a$ between the $\log_b$ and the $(x)$. Of course, this is not math; just mnemonics, and pretty weak at that.

Now, letting $b = e$ (so that $\log_b = \ln$) and $a = 10$, you get

$$\ln(x) = \log_{10}(x)\ln(10).$$

This last form is the one you need in this case. Just replace $\ln(x)$ in your expression with $\ln(10) \log_{10}(x)$. (Note that $\ln(10)$ is simply a constant, just like $10$ or $e$ or $\pi$, etc.)

The important bits to focus on, make sure you understand, and commit to memory, are the identities $x = a^{\log_a(x)}$ and $\log_r(s^t) = t \log_r(s)$. (For the last one, the special case where $t = 2$ is particularly easy to remember, and may help you remember the more general formula: $\log_r(s^2) = \log_r (s \cdot s) = \log_r(s) + \log_r(s) = 2\log_r(s).$)

  • $\begingroup$ so my equation should be? "y = 3101.420903 * log10(x) * 2.302585093 * - 8588.741253" although this doesnt seem to be giving me the numbers im looking for. I put my x y data points in here xuru.org/rt/LnR.asp to get the ln function of the curve, but im definitely doing something wrong. Im trying to figure out the equation for the curve (which is smooth) so I can find out y when x = n $\endgroup$
    – Ke.
    Mar 14, 2014 at 11:23
  • $\begingroup$ Yes, the equation should be as you have it, although I'd go ahead and multiply together the two constants you have there ($3101.420903$ and $2.302585093$). Also, the problem of getting $y$ when $x = n$, does not require converting to $\log_{10}$. All you need is to have an actual numeric (as opposed to symbolic) value for $x$, and then you just replace $x$ in the original equation with this actual value, take its natural logarithm, and off you go. $\endgroup$
    – kjo
    Mar 14, 2014 at 11:34
  • 1
    $\begingroup$ I think my data points were inaccurate so plotted them in this tool zizhujy.com/en-us/Plotter to get an exponential fit. Your simplication really helped me understand, thank you very much kjo $\endgroup$
    – Ke.
    Mar 14, 2014 at 12:09

Remember that: $$\ln x = \log_{e}x$$ Now just use the change of base formula, which is: $$\log_{a}x = \dfrac{\log x}{\log a}$$ We have our expression: $$3101.420903\ln x + 8588.741253$$ Step 1: Rewrite the equation. We can change $3101.420903\ln x + 8588.741253$ to $3101.420903\log_{e}x+8588.741253$.

Step 2: Apply the change of base formula. $$3101.420903\left(\dfrac{\log x}{\log e}\right)+8588.741253$$ $$\dfrac{3101.420903\log x}{\log e}+8588.741253$$ $$=\dfrac{3101.420903\log x}{\log e}+\dfrac{8588.741253\log e}{\log e}$$ $$=\dfrac{3101.420903\log x+8588.741253\log e}{\log e}$$ $$\displaystyle \boxed {\therefore 3101.420903\ln x+8588.741253=\dfrac{3191.420903\log x+8588.741253\log e}{\log e}}$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .