Changes

*<math>\ln(ab) = \ln(a)+\ln(b)</math> for all positive reals <math>a,b</math>.

*<math>\ln(ab) = \ln(a)+\ln(b)</math> for all positive reals <math>a,b</math>.

Proof: If <math>a,b > 0</math>, we can write <math>a = e^x</math> and <math>b= e^y</math>. It follows that <math>\ln(ab) = \ln(e^x e^y) = \ln(e^{x+y})</math>. By definition, <math>\ln(e^{x+y}) = x+y = \ln(e^x)+\ln(e^y)</math>. This last expression, of course, is <math>\ln(a)+\ln(b)</math>.

Proof: If <math>a,b > 0</math>, we can write <math>a = e^x</math> and <math>b= e^y</math>. It follows that <math>\ln(ab) = \ln(e^x e^y) = \ln(e^{x+y})</math>. By definition, <math>\ln(e^{x+y}) = x+y = \ln(e^x)+\ln(e^y)</math>. This last expression, of course, is <math>\ln(a)+\ln(b)</math>.

*<math>\ln(x^p) = p\ln(x)</math>

*<math>\ln(x^p) = p\ln(x)</math>

Proof: If <math>p</math> is a positive integer, this just follows from repeated application of the above-mentioned additive property of the logarithm. For <math>p = \frac{1}{n}</math>, note that the statement follows by observing that <math>\ln(x) = \ln((x^{1/n})^n) = n\ln(x^{1/n}</math>. Thus, <math>p\ln(x) = \ln(x^p)</math> for all rational numbers <math>p</math>. The statement must therefore hold for all reals <math>p</math> by continuity.

Proof: If <math>p</math> is a positive integer, this just follows from repeated application of the above-mentioned additive property of the logarithm. For <math>p = \frac{1}{n}</math>, note that the statement follows by observing that <math>\ln(x) = \ln((x^{1/n})^n) = n\ln(x^{1/n}</math>. Thus, <math>p\ln(x) = \ln(x^p)</math> for all rational numbers <math>p</math>. The statement must therefore hold for all reals <math>p</math> by continuity.

*<math>\ln(1) = 0</math>

*<math>\ln(1) = 0</math>

Proof: <math>\ln(1) = \ln(1\cdot 1) = \ln(1)+\ln(1)</math>, whence we must have <math>\ln(1)=0</math>.

Proof: <math>\ln(1) = \ln(1\cdot 1) = \ln(1)+\ln(1)</math>, whence we must have <math>\ln(1)=0</math>.

[[Category:Mathematics]]

[[Category:Mathematics]]