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