# Change base log formula?

Im trying to change the base log from ln to log with the following formula.

 y = a * ln⁡(x+c) + b


The ln equation is:

y = 17.686797186426052  * ln(x + 0.3) + -79.52230886298399


And considering that:

ln(X)=2.302585 log(X)


I was trying to change the equation to use log function (to this):

y = 17.686797186426052  * 2.302585 *  log(x + 0.3) + -79.52230886298399


However, for some reason it's not working. Id be really grateful for some pointers on where I might be going wrong?

Cheers

Rewrite the formula as \begin{align} y&=a\ln(x+c)+b\\ &=a\log_e(x+c)+b\\ &=a\frac{\log_k(x+c)}{\log_k e}+b\\ &=\frac{a}{\log_k e}\cdot\log_k(x+c)+b.\\ \end{align} If you want the logarithm in base-$10$, then you simply write $$y=\frac{a}{\log e}\cdot\log(x+c)+b.$$ Now, you can plug in the values of $a$, $b$, and $c$ without having trouble.

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• Hi thanks for the answer. Im wondering what "log e" means? is that log(2.71828), or log(x) * 2.71828 ?
– Ke.
Apr 7 '14 at 14:45
• @Ke. $\log_e x$ means the base of logarithm is $e$, the Euler's number and $e=2.71828...$. To understand what is base in the logarithm, take a look this one: $$\log_a x=b\;\quad\;\Rightarrow\;\quad x=a^b.$$ Similarly $$\log_e x=b\;\quad\;\Rightarrow\;\quad x=e^b.$$ Hence, $\log_e x\neq\log(2.71828)$ or $\log_e x\neq2.71828\log(x)$. Apr 7 '14 at 15:06
• Many thanks for your help :)
– Ke.
Apr 7 '14 at 16:30
• @Ke. You are welcome my friend. :) Apr 7 '14 at 16:32