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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

MT_
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Ke.
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2 Answers2

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Hint:

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

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

DonAntonio
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  • Still struggling im afraid. Im at a loss at how to apply this change of base to the equation Ive got. – Ke. Apr 06 '14 at 19:16
  • @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. – DonAntonio Apr 06 '14 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:

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

npisinp
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