There are a number of uses for logarithms, often abbreviated to logs, one is that it is the inverse function to raising a number to a power. Another is that it can convert between multipication and addition, for instance,
log 3 + log 4 = log 12
Logs
The 'log' of a number, to a given base, is the power to which the base must be raised to give the number.
e.g.,
if: log_{10}200 = 2.3010
then: 10^{2.3010}= 200
Consider the base 2
since 2^{3}=8 then log_{2}8 = 3
since 2^{4}=16 then log_{2}16 = 4
in general if the base = a and y=a^{x} then,
log_{a}y = x
Some Properties of logs
The log of the base itself is always unity | since a^{1} = a then log_{a}a = 1 |
The log of one is always zero | since a^{0} = 1 then log_{a}1 = 0 |
The log of a number is equal to minus the log of its reciprocal | log_{a}x = - log_{a}(1/x) |
log_{a}(x * y) = log_{a}x + log_{a}y | |
log_{a}(x ÷ y) = log_{a}x - log_{a}y | |
log_{a}(x)^{n} = n*log_{a}x | |
log_{a}^{n}√(x) = (1/n)*log_{a}x | |
log_{a}(x)^{n} = n*log_{a}x |
Examples
- ^{3}√(0.0838) = antilog(-0.389/3)
- (0.0273)^{2/3} = antilog(2/3 * log 0.0273) = 0.09067
- (6.023)^{-2.5} = antilog(-2.5 * log 6.023) = antilog(-2.5 * 0.7798) = antilog(-1.9495) = 0.01123
- (0.1276)^{-1.7} = antilog(-1.7 * log (0.1276) = antilog(-1.51997) = 33.11
Solution of exponential equations
An exponential equation is one in which the unknown quantity is an index.
Example 1:
3^{p} * 5^{(p-2)} = 8^{2p} * 7^{(1-p)}
p log 3 + (p-2) log 5 = 2p log 8 + (1-p) log 7
0.4771 p + 0.699(p-2) = 0.9031(2p) + 0.8451(1-p)
0.215 p = 2.2431
p=10.43
Example 2:
7.16^{y} = 1.92 ^{(y+2)}
y log 7.16 = (y+2)log1.92
0.8549 y = 0.2833 (y+2)
0.8549 y - 0.2833y = 0.5666
y = 0.5666/0.5716
Example 3:
200^{} = k * 12 ^{1.25}* 80 ^{-0.5}
log 200 = log k + 1.25*log12 +(-0.5 log 80)
2.301 = log k + 1.25*1.0792 - 0.5*1.9031
log k = 1.90355
k = 80.1
Bases other than 10
If 5 ^{2}= 25 then by definition log_{5}25 = 2
If 4 ^{3}= 64 then by definition log_{4}64 = 3
Examples
evaluate log_{6}216
let log_{6}216 = x
then 6 ^{x} = 216
x * log_{10}6 = log_{10}216
x = 2.3345/0.7782
x = 3
Logs to the power of 2
evaluate log_{2}1.87
let log_{2}1.87 = x
then 2 ^{x} = 1.87
x * log_{10}2 = log_{10}1.87
x = 0.2718/0.301
x = 0.9241
Natural or Naperian Logs
The base 'e' is very common base in engineering, where:
e = 1 + 1/1 + 1/(1*2) + 1/(1*2*3) + 1/(1*2*3*4) ...
the value of e to 5 decimal places is = 2.71828
To change a log from base e to base 10
log_{e}x = n
then e^{n} = x
n * log_{10}e = log_{10}x
n = log_{e}x = log_{10}x / log_{10}e
since 1 / log_{10}e = 2.3026
log_{e}x = log_{10}x * 2.3026