The function f(x) = a^x is a one to one function provided that a > 0 and a =! 1. Therefore f would have an inverse which we will call the logarithmic function.
What are logarithmic functions?
If a > 0 and a =! 1, then the function log_a(x), called the logarithm of x to the base a, is the inverse of the one to one function a^x:
Y = log_a(x) x = a^y, (a > 0, a =! 1)
Since a^x has the domain (-inf, inf), log_a(x) has the range (-inf, inf). Since a^x has the range (0,inf), log_a(x) has the domain (0, inf). Since a^x and log_a(x) are inverse functions, the following cancellation identities hold:
log_a (a^x) = x for all real x and a^(log_a x) = x for all x > 0.
The graphs of some examples of logarithmic functions are shown in the picture (a) below. They all pass through the point (0,1). Each graph is the reflection in the line y = x of the corresponding exponential graph in the picture (b).
From the laws of exponents we can derive the following laws of logarithmic functions.
Laws of logarithms: If x > 0, y > 0, a > 0, b > 0, a =! 1, b =! 1, then
(i) log_a(1) = 0
Proof: We know that a^0 = 1 for any real number a. Therefore by the definition of log we can say that log_a(1) = 0. Hence proved.
(ii) log_a(xy) = log_a (x) + log_a(y)
Proof: Let u = log_a(x) and v = log_a(y). By defining property of inverse of log function we know that x = a^u and y = a^v. Thus xy = a^u * a^v = a^(u+v). Inverting again we have, log_a(xy) = u + v = log_a(x) + log_a(y). Hence proved!
(iii) log_a(1/x) = - log_a(x)
(iv) log_a(x/y) = log_a(x) - log_a(y)
Proof: Let u = log_a(x) and v = log_a(y). Then again by defining property of inverse of log function we know that x = a^u and y = a^y. Thus a^u/a^v = a^(u-v). Inverting again we have, log_a(x/y) = u - v = log_a(x) - log_a(y). Hence proved. The rule number (iii) can be proved in the same manner.
(v) log_a(x^y) = y log_a(x)
(vi) log_a(x) = (log_b x)/( log_b a)
The logarithm law (vi) presented above shows that if you know the logarithms to a particular base b, you can calculate the logarithms to any other base a.
