# 10: Using logarithms - Log vs. Ln

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A common question exists regarding the use of logarithm base 10 ($$\log$$ or $$\log_{10}$$) vs. logarithm base $$e$$ ($$\ln$$). The logarithm base $$e$$ is called the natural logarithm since it arises from the integral:

$\ln (a) = \int_1^a \dfrac{dx}{x}\nonumber$

Of course, one can convert from $$\ln$$ to $$\log$$ with a constant multiplier.

$\ln (10^{\log a}) = \log (a) \ln(10) \approx 2.3025 \log (a)\nonumber$

but $$10^{\log a}=a$$ so

$\ln a \approx 2.4025 \log a\nonumber$

The analysis of the reaction order and rate constant using the method of initial rates is performed using the $$\log_{10}$$ function. This could have been done using the $$\ln$$ function just as well. The initial rate is given by

$r_o=k'[A]_0^a\nonumber$

The analysis can proceed by taking the logarithm base 10 of each side of the equation

$\log r_o = \log k' + a\log [A]_0\nonumber$

or the $$\ln$$ of each side of the equation

$\ln r_o = \ln k' + a\ln [A]_0\nonumber$

as long as one is consistent.

Once can think of the $$\log$$ or the $$\ln$$ as a way to 'linearize data' that has some kind of power law dependence. The only difference between these two functions is a scaling factor ($$\ln 10 \approx 2.3025$$) in the slope. Obviously, if you multiply both sides of the equation by the same number the relative values of the constants remains the same on both sides.