Calculating Logarithms: \log_a (b) = {\ln (b) \over \ln (a)}

[Gregg]

I read that

\log _2 (3) = {\ln (3) \over \ln (2)}

Is

\log _a (b) = {\ln (b) \over \ln (a)}

How?

 

[jgens]

Well, defining the logarithmic function as the inverse of the exponential function, you can prove the equality like this. Clearly,

b = a^{\log_a{(b)}}

Evaluating the logarithm base c of each side produces,

\log_c{(b)} = \log_a{(b)} \log_c{(a)}

Dividing through by \log_c{(a)} we get

\log_a{(b)} = \frac{\log_c{(b)}}{\log_c{(a)}}

As desired.

 

[zgozvrm]

(1) First of all, realize that \log_b(a) = x by definition, means that b^x = a

(2) We can show that \log_b(a^y) = y\log_b(a):

1st assume that \log_b(a^y) = x, then b^x = a^y, by (1). Now, we have (b^x)^\frac{1}{y} = (a^y)^\frac{1}{y}, or b^\frac{x}{y} = a. By definition of logarithms (1), this gives us \log_b(a) = \frac{x}{y}, and finally y\log_b(a) = x


Now, given \log_a(b) = x, we have:

b = a^x, by (1)

\ln(b) = \ln(a^x)

\ln(b) = x\ln(a), by (2) (remember that \ln a = \log_e(a))

and, finally x = \frac{\ln(b)}{\ln(a)}

or, more generally, it can be shown that

\log_a(b) = \frac{\log_x(b)}{\log_x(a)}, for any positive value of x