>>11948444Anon:
Think of log base 10. 10^-1= 0.1, 10^0=1, 10^1=10, 10^2 = 100, etc, so log10(.1)=-1, log10(1)=0, log10(10)=1, log10(100)=2, etc. For numbers between this, your calculator can figure out the exact exponent. eg log10(50) is about 1.699 because 10^1.699...=50; you can't easily calculate that yourself, but since 10 < 50 < 100, you can at least know log10(50) is between 1 and 2.
Natural logarithms are just base e = 2.718... . So ln(1)=0, ln (2.718...)=1, ln(7.389...)=2, ln(20.085...)=3, etc. Again, calculating exactly is tough, but at this point it's all about symbolic manipulation.
On that note, logarithms have really convenient math properties. As useful practice, I'd encourage you to try proving these yourself -- they're helpful for understanding and not too hard.
ln(e^x) = e^ln(x) = x. (True for most, but not all, numbers, at least without getting into complex analysis. Think about which ones may not work here by looking at their graphs and seeing where each function is defined.)
log base a (x) = ln(x) / ln (a). Note that you don't need ln on the right side, log x over log a in any base yields log x base a. Try proving by putting both sides e^(...) and applying exponent rules.
ln(x^y) = y*ln(x). This is a crucial property for proofs. Also, again, you don't need ln (aka log base e) here -- all of these properties are general properties of logarithms.
ln(xy) = ln(x) + ln(y). This is where slide rules came from: before calculators, really big, precise multiplications could be done easily by taking the log of the two inputs, adding them, and de-converting, using pre-calculated logarithms marked out on the slide rule.
ln(1/x) = -ln(x). Follows from prior rules, try proving.
Now, you may be wondering: why this 'e'? Honestly, the answer takes calculus (or at least a good answer, anyways, IMO). Teaser: d/dx a^x = a^x * ln(a). Also e^(i*pi) + 1 = 0, and a million other important identities
