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

[Gregg]

I read that

[tex] \log _2 (3) = {\ln (3) \over \ln (2)} [/tex]

Is

[tex] \log _a (b) = {\ln (b) \over \ln (a)} [/tex]

How?

 

[jgens]

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

[tex]b = a^{\log_a{(b)}}[/tex]

Evaluating the logarithm base [itex]c[/itex] of each side produces,

[tex]\log_c{(b)} = \log_a{(b)} \log_c{(a)}[/tex]

Dividing through by [itex]\log_c{(a)}[/itex] we get

[tex]\log_a{(b)} = \frac{\log_c{(b)}}{\log_c{(a)}}[/tex]

As desired.

 

[zgozvrm]

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

(2) We can show that [tex]\log_b(a^y) = y\log_b(a)[/tex]:

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


Now, given [tex]\log_a(b) = x[/tex], we have:

[tex]b = a^x[/tex], by (1)

[tex]\ln(b) = \ln(a^x)[/tex]

[tex]\ln(b) = x\ln(a)[/tex], by (2) (remember that [tex]\ln a = \log_e(a)[/tex])

and, finally [tex] x = \frac{\ln(b)}{\ln(a)} [/tex]

or, more generally, it can be shown that

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