Definition of the Logarithm Function

Question

I am currently reading Spivak's Calculus. I have an older version. In chapter 17 the author presents a definition of the logarithm function as $\log x=\int_1^x \frac{1}{t}dt$.

He begins in searching the derivative of a function which satisfies $f(x+y) = f(x) \cdot f(y)$, finds $f'(x) = \alpha \cdot f(x)$. Using the inverse function $f^{-1} = \log_{10}$ he arrives at $\log_{10}' = \frac{1}{f(f^{-1}(x))} = \frac{1}{\alpha x}$. Since $\log_{10}$ is the anti-derivative of $\frac{1}{\alpha x}$ he uses the second fundamental theorem of calculus to do the following integration:

$$\frac{1}{\alpha} \int_1^x \frac{1}{t}dt = \log_{10}x - \log_{10}1 = \log_{10}x$$

Now my question is, why does the factor $\frac{1}{\alpha}$ vanish in this equation?

I hope that I have provided enough intermediate steps to give some context. The whole argument is rather lengthy to present here. If otherwise, please comment.

Answer 1

Now my question is, why does the factor $\frac1α$ vanish in this equation?

The factor does not "vanish". The choice of $\alpha = \ln 10$ resp. of using $\log_{10}$ is not very natural from a mathematical standpoint. Anyways, in the equation

$$\frac{1}{\alpha} \int_1^x \frac{1}{t}dt = \log_{10}x - \log_{10} = \log_{10}x$$ the integral is just the definition of the Natural Logarithm $\ln$, and using that to rewrite the equation we get:

$$\frac{1}{\alpha} \ln x = \frac{\ln x}{\ln 10} - \frac{\ln 1}{\ln 10} = \frac{\ln x}{\ln 10} = \log_{10}x$$

This is just the rule how to transform logs in different bases, namely $$\log_b a = \frac{\log_c a}{\log_c b}$$ for $a, b, c > 0$ and $b,c\neq 1$.

Answer 2

The $\frac{1}{\alpha}$ does not "vanish"... Using $\log_{10}'(t) = \frac{1}{\alpha t}$, we have $$ \frac{1}{\alpha} \int_1^x \frac{1}{t}dt = \int_1^x \frac{1}{\alpha t}dt= \log_{10}x - \log_{10}1 = \log_{10}x $$