Properties of Logarithms?

Question

Use properties of logarithms to write each logarithmic in terms of a and b if In2 = a and In3 = b.

(1) In(2/3)

(2) In 0.5

Answer 1

There are only 3 properties of logs and you have to know them. They come from the properties of exponents, because a log IS an exponent.

log A + log B = log (AB)
log A - log B = log (A/B)
log A^p = p log A

also, use the definition (a lot): log_b A = c means b^c = A.

#1) ln (2/3) = ln 2 - ln 3 = a - b

#2) ln 0.5 = ln (1/2) = ln 1 - ln 2 = 0 - a = - a

note that ln 1 is 0 because e^0 = 1.

that's it! :)

Answer 2

If two numbers are multiplied, their logarithms are added.
If two numbers are divided, their logarithms are subtracted.

ln(2/3) = ln(2) - ln(3) = a - b

ln(0.5) = ln(1/2) = ln(1) - ln(2) = 0 - ln(2) = -a

The rule comes from how exponents are used. If the base is c, and we need to multiply c^3 by c^4, we could do it the long way:

(c*c*c)*(c*c*c*c) = c*c*c*c*c*c*c = c^7
or use the shortcut: (c^3)*(c^4) = c^(3+4) = c^7
in multiplying numbers that are all written in the same base, we add the exponents.

Division;

c^7 / c^5 =
(c*c*c*c*c*c*c) / (c*c*c*c*c) =
(c*c)* (c*c*c*c*c)/(c*c*c*c*c) = (c*c)*1 = c^2
or
(c^7)/(c^5) = c^(7-5) = c^2

since any (non-zero) base raise to the power 0 is 1
c^0 = 1 for any c>0
then log(1)=0 in any base
(including ln(1) - 0)

therefore when you have a power as a denominator, you can replace it with the negative power as a numerator:

1/c^4 = (c^0)/(c^4) = c^(0-4) = c^-4 (c to the power -4)

with logs (log_c(x) means the log of x in base c)
log_c(1/c^4) = log_c(c^-4) = -4

Answer 3

ln(2/3) = ln 2 - ln 3 = a - b.
ln 0.5 = ln 1/2 = ln 1 - ln 2 = -ln 2 = -a.

Answer 4

ln(a/b)=ln(a)-ln(b)
ln(2/3)=ln(2)-ln(3)=a-b

ln(1/p)= - ln(p)
ln(0.5)=ln(1/2)= - ln(2)= - a.