Important Basics on exponentials and logarithms

The number e is 2.718…, you get it by \((1 + \frac{1}{n}))^n\) for very high n. See (http://www.purplemath.com/modules/expofcns5.htm). Or in R simply exp(1) = 2.7182818
log and exp: log(exp(1)) is 1, log(exp(11)) is 11, so log in R is the log to the basis e
to understand exponential, get aware that independent on the base, to the power of 0 is always 1 (https://www.khanacademy.org/math/pre-algebra/exponents-radicals/World-of-exponents/v/raising-a-number-to-the-0th-and-1st-power):
exp(2) = 1 * e * e
exp(1) = 1 * e
exp(0) = 1
another way to see why 2^0, 10^0 and e^0 are all 1, is: to get from 2^3 to 2^2 you divide by 2. From 2^2 to 2^1 you divide by 2. Same from 2^1 to 2^0 you divide by 2, and 2 divided by 2 is 1.
so see in the graph all exponentials meet at the point 0,1

x <- seq(from = -2, to = 2, by = 0.1)
y <- exp(x)
y1 <- 2^x
y2 <- 10^x
df <- data.frame(x = x, exp = y, base_2 = y1, base_10 = y2)
df <- gather(df, Base, Value, -x)

ggplot(df, aes(x=x, y = Value, colour = Base)) + geom_vline(xintercept = 0, color = cbPalette[1]) + geom_line() + scale_colour_manual(values = cbPalette[2:4]) + ylim(0,5) + theme_bw(14) + ylab("Exponential") + xlab("'log'") + theme(panel.grid.minor = element_blank()) + theme(panel.background = element_rect(colour = cbPalette[1],size = 2))

So \(base^{0}\) is always 1, \(base^{1} = base\)

Some exponential rules

First rule: \(2^{2}*2^{3} = 2^{2+3}\)
Second rule: \(\frac{2^{3}}{2^{2}} = 2^{3-2}\)
- consequently: \(2^{-3} = 2^{0-3} = \frac{2^{0}}{2^{3}} = \frac{1}{2^{3}}\)
so exp(x) is never negative (unless the base is negative). For x from -Inf to 0, exp(x) ranges from 0 to 1. exp(1) is the base. For integers x > 1 exponentials are easy to understand. But what for x between 0 and 1. You know that exp(x) in that case will be between 1 and base. But why.
see (https://www.mathsisfun.com/algebra/exponent-fractional.html)
- e.g. \(2^{\frac{1}{3}}\). Thanks to the first rule you know: \(2^{\frac{1}{3}}\times2^{\frac{1}{3}}\times2^{\frac{1}{3}} = 2^{1} = 2\) That means that \(2^{\frac{1}{3}} = \sqrt[3]{2}\)
  - so albeit that does not help much with imagining the calculation, but remember \(x^{\frac{1}{4}}\) is a number that multiplied for times with itself is x.
and more complicated fractions? \(2^{\frac{3}{4}} = 2^{3*\frac{1}{4}}\)
Third rule: \((2^{3})^{3} = 2^{3*3} = 2^{9}\)
- so \(2^{\frac{3}{4}} = 2^{3*\frac{1}{4}} = \sqrt[4]{2^{3}}\)

And for the logarithms

log of y to a base is the number x you have to power base with to get y.
so in the plot above, choose a y value, look what is the corresponding x value, it is the log(y) to the respective base (in the plot exp, 2, 10)
note for relative abundances, y is betwen 0 and 1, so the log(y) are from -Inf to 0.
- log(1) is 0 independent of base
What if you have log to base e (as in R), but you want to base 2?
- \(2^{3} = (e^{z})^3 = e^{z*3}\) with \(e^{z} = 2\) so \(z = log(2)\) so the log base e of 2.
  - that means \(log_{2}(y)\) is \(z\times log(y) = \frac{log(y)}{log(2)}\)
so general rule: log of y to the base x = log(y)/log(x)

all other rules come directly from the exponential rules:

take \(a = 2^{4}\) so log(a) = 4 (base 2) and \(b = 2^{3}\), so log(b) = 2
then rule 1: log(a*b) = log(a) + log(b) (think of base 2, \(2^{4}\times2^{3} = 2^{3+4}\) )
rule 2: \(log(\frac{a}{b}) = log(a) - log(b)\)
rule 3: \(log(a^{z}) = z*log(a)\)

x <- seq(from = 0, to = 1, by = 0.01)
y <- log(x)
df <- data.frame(x = x, log_x = y)
df$Base = "exp"

ggplot(df, aes(x=x, y = log_x, colour = Base)) + geom_vline(xintercept = 0, color = cbPalette[1]) + geom_line() + scale_colour_manual(values = cbPalette[2]) +  theme_bw(14) + ylab("log(x)") + theme(panel.grid.minor = element_blank()) + theme(panel.background = element_rect(colour = cbPalette[1],size = 2))

Logarithm

Thorsten Brach

15 May 2015

Important Basics on exponentials and logarithms

And for the logarithms