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Im wondering how I could go about substituting $\log_{10}$ for $\ln$ in the following formula?

$y=a+b\ln(x+c)$

Is there a simple way of doing this?

Cheers

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  • $\begingroup$ Do you mean substitute log10 for ln? Or is did you typo the equation? $\endgroup$ Apr 6, 2014 at 19:10

2 Answers 2

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Hint:

For any $\;1\neq a,b>0\;$ :

$$\log_ax=\frac{\log_bx}{\log_ba}$$

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  • $\begingroup$ Still struggling im afraid. Im at a loss at how to apply this change of base to the equation Ive got. $\endgroup$
    – Ke.
    Apr 6, 2014 at 19:16
  • $\begingroup$ @Ke., your function is not clear at all. I can't help you with that, but I think the above is pretty clear, isn't it? Of course, you need to read about logarithms a little. The above's called "property of change of basis in logarithms". You can even google it. $\endgroup$
    – DonAntonio
    Apr 6, 2014 at 19:19
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There is a a formula to change from base $k$ to base $b$. See $\log_b x=\dfrac{\log_k x}{\log_k b}.$

In your equation you have $\log(x+c)$ ($\log$ is the natural logarithm). If we apply the change of base rule we get:

$$\log(x+c)=\dfrac{\log_{10}(x+c)}{\log_{10} e},$$

where $e=\exp(1)$ is the exponetial of $1$ $e=2.71828$.

Therefore your equation is simply:

$$y=a+\dfrac{b}{\log_{10} e}\log_{10}(x+c),$$

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  • $\begingroup$ I sort of get how to change base, but just not sure how to apply this to the equation I have :/ $\endgroup$
    – Ke.
    Apr 6, 2014 at 19:19
  • $\begingroup$ Ok. I edited my answer. $\endgroup$
    – npisinp
    Apr 6, 2014 at 19:27

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