How to convert the ln part of this equation to log10?

Question

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.

Cheers.

Answer 1

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 $b=e$.

Answer 2

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: $\log_r(s^2) = \log_r (s \cdot s) = \log_r(s) + \log_r(s) = 2\log_r(s).$)

Answer 3

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. $$3101.420903\left(\dfrac{\log x}{\log e}\right)+8588.741253$$ $$\dfrac{3101.420903\log x}{\log e}+8588.741253$$ $$=\dfrac{3101.420903\log x}{\log e}+\dfrac{8588.741253\log e}{\log e}$$ $$=\dfrac{3101.420903\log x+8588.741253\log e}{\log e}$$ $$\displaystyle \boxed {\therefore 3101.420903\ln x+8588.741253=\dfrac{3191.420903\log x+8588.741253\log e}{\log e}}$$