Calculus proof of ln(ab)= lna + lnb

Question

My calculus book states the following theorem of the properties of natural logarithms:

If a, b > 0 , then ln(ab)= lna + lnb

The author goes on to prove this theorem as follows

I do not understand what property allowed the author to use the substitution U = t/a because the original variable in the second integral is "t" and clearly U is not the same as t. Shouldn't U = t.

Answer 1

When you have a definite integral, the variable which you are integrating with respect to is a "dummy variable": in the sense that it does not matter what you call it. Thus, $$\int_a^b\frac1tdt,\;\int_a^b\frac1udu,\;\int_a^b\frac1sds,\;\int_a^b\frac1\zeta d\zeta$$ all mean exactly the same thing and have exactly the same value. After the substitution $u=t/a$, we obtain $$\int_a^{ab}\frac1tdt=\int_1^b\frac1udu=\ln b.$$

Answer 2

Here is an alternative approach.

Since $\ln(x) = \int_1^x \frac1t$, we have $\ln'(x)=\frac1x$.

Let $ f(x)=\ln(ax)- \ln(a) - \ln(x) $. Then $f'(x)=0$ and $f(1)=0$ imply $f(x)=0$ for all $x$.