3. The Logarithm Laws
by M. Bourne
Since a logarithm is simply an exponent which is just being written down on the line, we expect the logarithm laws to work the same as the rules for exponents, and luckily, they do.
| Exponents | Logarithms |
|---|---|
| `b^m × b^n = ` `b^(m+n)` | `log_b xy = ` ` log_b x + log_b y` |
| `b^m ÷ b^n = ` `b^(m-n)` | `log_b (x/y) = ` `log_b x − log_b y` |
| `(b^m)^n = b^(mn)` | `log_b (x^n) =` ` n log_b x` |
| `b^1 = b` | `log_b (b) = 1` |
| `b^0 = 1` | `log_b (1) = 0` |
Note: On our calculators, "log" (without any base) is taken to mean "log base 10". So, for example "log 7" means "log107".
Examples
1. Expand
log 7x
as the sum of 2 logarithms.
Answer
Using the first law given above, our answer is
`log 7x = log 7 + log x`
Note 1: This has the same meaning as `10^7 xx 10^x = 10^(7+x)`
Note 2: This question is not the same as `log_7 x`, which means "log of x to the base `7`", which is quite different.
2. Using your calculator, show that
`log (20/5) = log 20 − log 5`.
Answer
I am using numbers this time so you can convince yourself that the log law works.
LHS
`= log (20/5)`
`= log 4 `
`= 0.60206` (using calculator)
Now
RHS
`= log 20 − log 5`
`= 1.30103 − 0.69897` (using calculator)
`= 0.60206 `
= LHS
We have shown that the second logaritm law above works for our number example.
3. Express as a multiple of logarithms: log x5.
Answer
Using the third logarithm law, we have
`log x^5 = 5 log x`
We have expressed it as a multiple of a logarithm, and it no longer involves an exponent.
Note 1: Each of the following is equal to 1:
log6 6 = log10 10 = logx x = loga a = 1
The equivalent statements, using ordinary exponents, are as follows:
61 = 6
101 = 10
x1 = x
a1 = a
Note 2: All of the following are equivalent to `0`:
log7 1 = log10 1 = loge1 = logx 1 = 0
The equivalent statments in exponential form are:
70 = 1
100 = 1
e0 = 1
x0 = 1
Exercises
1. Express as a sum, difference, or multiple of logarithms:
`log_3((root(3)y)/8)`
Answer
`log_3((root(3)y)/8)`
` =log_3(root(3)y)-log_3(8)`
`=log_3(y^(1//3))-log_3(2^3)`
`=1/3log_3(y)-3 log_3(2)`
2. Express
2 loge 2 + 3 loge n
as the logarithm of a single quantity.
Answer
Applying the logarithm laws, we have:
2 loge 2 + 3 loge n
= loge 4 + loge n3
= loge 4n3
Note: The logarithm to base e is a very important logarithm. You will meet it first in Natural Logs (Base e) and will see it throughout the calculus chapters later.
3. Determine the exact value of:
`log_3root(4)27`
Answer
Let
`log_3root(4)27=x`
Then
`3^x=root(4)27`
Now
`root(4)27=root(4)(3^3)=3^(3//4)`
So `x = 3/4`.
Therefore
`log_3root(4)27=3/4`
4. Solve for y in terms of x:
log2 x + log2 y = 1
Answer
Using the first log law, we can write:
log2 xy = 1
Then xy = 21
So
`y=2/x`