As we move into the next article we will be dealing with the pH scale which is a log scale. We can use this fact to our advantage when solving questions and for that reason, we will take a quick break from talking about acids and bases explicitly.

A logarithm or a log is an exponent retriever. Logs can have different bases and each specifies the type of exponents we can retrieve. For example, log_{10} allows us to retrieve the exponent for 10^{x} while log 2 will allow us to retrieve any exponent that is 2^{x}. So if I take log_{10}(10^{6}) then the answer to this mathematical expression would be 6. Hence the log retrieved the exponent as an answer.

On the MCAT we will only be expected to deal with two specific log bases 10 and *e*. From here on out when I write log I am actually writing log_{10} however, this is the standard log value so we don’t have to show the 10. Log* _{e}* is also called the natural logarithm and is represented with

It’s great to have a definition of this stuff but how do we actually use this information? There are two major uses: conceptually to answer questions and in calculations. We will dive into the conceptual understanding first since I find that we rarely need to calculate log values and it is usually very easy to do so.

In order to understand logs from a conceptual point of view, we must first discuss log scales and why they are used. Scientists realized that visualizing data that differ by multiple orders of magnitude was only possible with exceedingly large pieces of paper or cramped scales that fail to show the meaningful differences between points.

Instead of printing their graphs on billboard size pieces of paper, they opted to convert the scale itself into a log scale. This compresses the scale allowing meaningful difference to be shown in a reasonable fashion. However, looks can be deceiving.

On a normal log scale, a change of 1 unit is equal to a times 10-factor change. For instance, if we went from a value of 3 to 6 on a log scale the true difference between the two points is 10^{3} or a 1000 unit difference, not a 3 unit difference. Why? Each unit stands in for a 10 times difference and we went up 3 units. This is important because we choose an answer choice on the log scale we should never ever pick a number that isn’t divisible by 10 unless we moved by a decimal amount on our scale.

The one caveat to this is the decibel scale for measuring the intensity of sounds. Instead of being a typical log scale where 1 unit corresponds to a 10 times factor difference, it is a 10-log scale. This means that for every 10 units we move up or down on the scale we move one factor of 10. So a shift in 30 units on this scale corresponds to a 10^{3} or a 1000 unit difference. Either way, we should still pick an answer that corresponds to a factor of 10.

On occasion, we will need to calculate a log value. In most cases, these are very simple. For example pH = -log[H+] . So if we have an H^{+} concentration of 10^{-6} M the pH would be 6. Since we first retrieve the exponent of 10^{-6} or -6 then multiply it by -1 we arrive at a pH of 6 for an H^{+} concentration of 10^{-6} M.

What if we need to go the other direction. When undoing the log function we raise 10 to the retrieved exponent performing any necessary calculations to our exponent value prior. For example, if when wanted to know what hydrogen concentration corresponded to a pH of 6 we would begin by tacking on a negative sign. We do this first because the negative sign is outside of the log value so we need to carry out any of those functions prior to raising to our exponent. From here we can go ahead raise 10 to the -6. So a pH of 6 means there is a 10^{-6} M H^{+} concentration.

All of that is great when dealing with simple 1×10^{x} values but what about other situations where the power of 10 is multiplied by a number other than 1. Say 2×10^{-6}. Then what? First, we have to realize that the AAMC hardly if ever requires you to calculate log values of this form. Instead, they will almost always ask you about 10^{x} values or space the answer choices out far enough that it is pointless to calculate an exact answer.

I mention this because you will always want to double check and see if you even need to get to an exact answer. If you do then we can use the following trick

[latexpage]

\[log (A \times 10^B) = B + 0.A\]

For example:

\[log (2 \times 10^{6}) = 6 + 0.2 = 6.3\]

For a pH value this changes a bit and the formula becomes due to the negative sign in front of the log value:

\[-log (A \times 10^B) = -B – 0.A\]

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