Bevis

Question 1

Logaritmlagarna

Answer 1

Alla tre bevis bygger på att f(x)=e^x är injektiv:

e^VL = e^HL ⇔ VL=HL

1. ln(xy) = ln(x)+ ln(y)
   e^VL=e^ln(xy) = xy = e^ln(x) • e^ln(y) = e^ln(x)+ln(y) = HL
   ∴ ln(xy) = ln(x) + ln(y)
2. ln(x/y) = ln(x) - ln(y)
   e^VL=e^ln(x/y) = x/y = (e^ln(x))/(e^ln(y)) = e^ln(x)-ln(y)= e^HL
   ∴ ln(x/y) = ln(x)-ln(y)
3. ln(x^p) = p•ln(x)
   e^VL=e^ln(x^p) = x^p = (e^ln(x))^p = e^p•ln(x) = HL
   ∴ ln(x^p) = p•ln(x)

VSV

Question 2

Log↓a(x)= ln x / ln y

Answer 2

log↓a(x) <=> a^y=x
<=> ylna = ln x
<=> y= ln x / ln a
alltså log↓a(x) = ln x/lny
vsv