Integral Calculus, Adding Logs

Adding Logs

Given two positive reals a and b, the log of a/b is the integral of 1/t as t runs from 1 to a/b. Integrate by substitution, replacing t with u/b. This gives the integral of 1/u from b to a, or log(a)-log(b). Verify that log(1/b) = -log(b). Do this by substitution; replace x with 1/u. Then replace b with 1/b in log(a/b), and show that log(a×b) = log(a)+log(b).

If you know about continuous groups, the log function is a continuous group homomorphism from the positive reals under multiplication to the reals under addition. In fact it is a group isomorphism, since the map is reversible via the exponential function. As groups, or even metric spaces, the positive reals under multiplication are indistinguishable from the reals under addition.

If a is positive and b is real, define a^b as exp(b×log(a)). Verify a¹ = a, and a⁰ = 1. Since n×log(a) is log(a) added to itself n times, equals log(a×a×…×a), we see that aⁿ agrees with a to the n^th power, in the traditional sense. since -log(x) = log(1/x), a^-n is indeed the inverse of aⁿ. The analytic definition agrees with the traditional definition even for negative exponents. But what about rational and real exponents? First let's prove some properties of exponentiation.

Verify that a^b×a^c = a^b+c. Take logs of both sides, remembering that log(xy) = log(x)+log(y), and get b+c = b+c. Now let c = -b and we have a^b×a^-b = a^b-b = a⁰ = 1. Thus a^-b is the inverse of a^b. Finally, a^bc = exp(bc×log(a)). Group b with log(a), or c with log(a), as you prefer. If b is folded into log(a), this formula becomes x^c, where the log of x is b times the log of a. This means x = a^b, and a to the bc is a to the b to the c. By symmetry, a to the bc is also a to the c to the b. These properties hold for all real exponents b and c, and a > 0.

Verify that a^1/n is the n^th root of a. This because a to the 1/n to the n is a¹ = a. In other words, a to the 1/n is that number x such that xⁿ = a. This agrees with the traditional definition of a to the 1/n. Combine these results to describe rational exponents. That is, a^p/q is a to the p to the 1/q, is the qth root of a^p, as expected. The two formulations of a^b, one from iterative multiplication and the other from integral calculus, agree on all rational exponents. Since they are both continuous functions, they agree on all real exponents. We have simply defined exponentiation in another way.

We expect the derivative of x^b to be bx^b-1, just as it was before. Verify this using our new exponential formula and the chain rule. In addition, we can now differentiate a^x, giving log(a)×a^x.

Let E = exp(1). Now we see that exp(x) and E^x are synonymous. Sometimes the exponential notation E^x is clearer. Exp(x) = E^x is the irrational number E raised to the x power. E is approximately 2.71828.