Logs can also be negative. In fact, the log of a number less than 1 must be negative: Since 0.01 = 10^{-2}, we must have log 0.01 = -2. Also, log 0.001 = -3.

Since 1 = 10^{0}, we have log 1 = 0.

For numbers that are not exact powers of 10 (positive or negative), we have to resort to a calculator (or tables in the days before calculators) to find the log. We won't go into how log tables were originally constructed (centuries ago), or how calculators manage to find logs. But in using a calculator it is wise to make note of whether your answer is coming out reasonably -- as we will discuss below.

[Also make sure that you are using log to the base 10 on the calculator. Calculators usually also have logs to the base e, which we won't discuss here, or use.]

log 1 = 0 and log 10 = 1. Therefore we expect that if n is a number between 1 and 10, log n should be a positive number between 0 and 1. Use your calculator to verify the following:

- log 2 = 0.301
- log 5 = 0.699
- log 9 = 0.954

Notice that log 9 is close to 1.00, as it should be, since log 10 = 1.00. It might be useful to choose a few more points between 1 and 10, and plot log x vs. x.

The square root of 10 is 3.162 (as you can verify). What should be log of 3.162 be? Looking at the results above for log 2 and log 5, you should expect the answer to be between .3 and .7. In fact, the answer is exactly 0.5 (except for a small round-off error). Why?

The important point about logarithms is that they convert products to sums. For two numbers, a and b,

log(ab) = log a + log b | Equ. (1) |

As an example, let a = 2, b = 5. Verify from the short list above, that log 2 + log 5 = 1.000, which is the log of 10. [log 10 = log 2 + log 5]

The general proof of Equ. (1) is given on this link.

Equ. (1) can be generalized to:

log(a X b X c X ...) = log a + log b + log c + ... | Equ. (2) |

As a special case, if the product is a X a X a ..., or a^{m}, for some integer, m, then

log (a^{m}) = log a + log a + etc. (m times), or | |

log (a^{m}) = mlog a | Equ. (3) |

Finally, we can generalize to the case where the exponent need not be an integer, in fact need not even be a positive number:

log (a^{x}) = xlog a. | Equ. (4) |

Since the log is the inverse of the exponential, we can use logs to solve equations in which the unknown is in the exponent. [This is analogous to using division to solve an equation when the unknown is in a product -- since division is the inverse operation of multiplication: e.g. 5x = 30, we divide to find x.]

If we have an equation like:

we simply take the log of both sides: Using Equ. (4), the log of the left side is y times the log of 6:

log 6 (from my calculator) is 0.778. On the right side we have log 1,350 = 3.13. [Note that log 1,000 = 3, log 10,000 = 4, so we expect log 1,350 to be between 3 and 4, probably close to 3.]

Our equation is now:

y = 3.13/0.778 = 4.02

This happens to come out close to an integer. In fact, 6^{4} = 1,296, so the answer should be close to 4.00.

Other examples:

x(-0.699) = -2.30

x = 3.29

How about

Here we don't use the calculator to find the log of the right side. It is -8.

n(-0.921) = -8

n = 8.69

Since 0.12 is approximately 1/10, we expect that we have to raise it to about the 8th. power to get 10^{-8}. And that's how it comes out. We get the power 8.69.

More examples which you can verify:

x = 5.73

9.2^{x} = 10^{6}

x = 6.23

0.15^{x} = 10^{-8}

x = 9.71

0.15^{x} = 10^{6}

x = - 7.28

Answer to earlier question: Why is the log of 3.162 (the square root of 10) equal to 0.5? The square root of 10 is 10^{1/2} = 10^{0.5}, and so the log is 0.5.