Understanding logarithms

A popular topic on OSC revision courses is logarithms. I usually start by asking who knows what a logarithm is, to be met by most of the class staring at their fingernails! To fully comprehend the laws of logarithms, and how to use them, it really helps to start with a clear understanding of the basics: and as long as you can cope with indices (powers), logs follow closely behind.

A logarithm is really just a power. If I asked you: “What power of 2 gives you 32?”, I hope you would be able to tell me the answer: 5. The same question could have been asked as: “What is the log of 32 to the base 2?” Answer: 5. In other words, the log of a number is simply the power to which you must raise the base to get the number. The statements: 25 = 32 and log232 = 5 are exactly the same, just as 72 = 49 and √49 = 7 are equivalent statements. Taking the square root is the inverse of squaring, and taking the log is the inverse of raising to a power.

Since you must understand negative and fractional indices as well as positive integer indices, check that you understand these log statements:

{\log }_{10}}1000 = 3}
{\log }_{9}}3 = \frac{1}{2}}
{\log }_{4}}\frac{1}{16} = -2}

In other words, never forget the that the statement logbx = y is exactly the same as xby, and you’ll find this relevant when trying to solve equations which have logs in them – there’s an example a bit further on.

The three laws of logarithms are derived from the laws of indices such as
xa × xb = xa + b. It’s essential to learn them, and use them correctly. The laws apply to logarithms to any base, but within each law logs must be to the same base. They are:

\log a + \log b = \log (ab)
\log a - \log b = \log (\frac{a}{b})
\log {a^n} = n\log a

The last one is particularly useful when solving equations where the unknown is a power. For example:

{3^{(x + 1)}} = 50 \newline \log {3^{(x + 1)}} = \log 50 \newline (x + 1)\log 3 = \log 50 \newline x + 1 = {{\log 50} \over {\log 3}} \newline x + 1 = 3.561 \newline x = 2.561

Note that the calculation can use logs to any base – use log or ln on your calculator.

Now, how about using the laws to solve this equation:  log2x + log24 = 5? Using the first law, the left hand side will simplify to log24x. Then, using the “never forget” bit I mentioned earlier, we can turn log24x = 5 into 4x = 25, from which we can easily work out that x = 8.

An alternative method is to ensure every term in the equation is written as a log to base 2. So, since 5 = log232, the equation becomes:

log2x + log24 = log232
log2x = log232 – log24
log2xlog2(using the second law)

So x = 8

Finally, you may occasionally have to use the “change of base” rule to convert logs from one base into another one. This states that {\log _b}x = {{{{\log }_a}x} \over {{{\log }_a}b}}.

For example, use logs base 10 to find out log458:

{\log _4}58 = {{{{\log }_{10}}58} \over {{{\log }_{10}}4}} = 2.929

You’ll find more detail, and plenty of examples and practice, in my revision guides. In both the HL guide and the SL guide I cover manipulating logarithms, as well as log functions.

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