# Logarithm Laws

If you are studying logarithms it is reasonably safe to assume that you’re already reasonably familiar with index rules; those shortcuts that allow for swift calculation of exponents when dealing with equal bases. The log laws are really just a rearrangement of these.

First, multiplying. Since when you multiply terms with the same base you add the indices, the log of a product is equal to the sum of the logs of the factors of that product. This is best demonstrated starting with the index rule and working through an example.

\begin {aligned}a^m \times a^n &= a^{m+n} \\ 8 \times 16 &= 128 \\2^3 \times 2^4 &= 2^7 \\ \log_2{8} + \log_2{16} &= \log_2{128} \\ \log_a{m} + \log_a{n} &= \log_a{mn}\end{aligned}Index / log law for multiplication / addition

Second, division. As the inverse of multiplication this one is now hopefully easy to spot: subtracting the parts. The log of a quotient is equal to the difference between the logs of the dividend and the divisor (numerator and denominator).

\begin {aligned}\frac{a^m}{a^n} &= a^{m-n} \\ \log{\frac{m}{n}}&= \log{m} - \log{n}\end{aligned}Index / log law for division / subtraction

Using this rule for division it follows that we should be able to find the value of a logarithm of 1. As any value divided by itself is 1, choosing a value expressed in index form allows us to apply the previous rule.

\begin {aligned}\frac{a^b}{a^b} &= 1 \\ a^{b-b} &= 1 \\ \log_a{1} &= b-b \\ \log_a{1} &= 0 \end {aligned}The log of 1 in any base is 0

Terms with indices raised to a power are an extension of the multiplication of indices already described. This approach can be applied to the addition of logarithms:

\begin {aligned}(a^m)^n &= a^m \times a^m \times \ldots \times a^m \\ \log_a{m^n} &= \log_a{m} + \log_a{m} + \ldots + \log_a{m} \\ \log_a{m^n} &= n \log_a{m}\end {aligned}Powers in logarithms

# Calculations with the Normal Distribution

When should you divide by n, or use n/n-1 with the variance when calculating values based on normally distributed variables?

Question 3 of OCR’s 2014 S2 unit is this:
As we know that G is normally distributed and we have a frequency distribution it is reasonable to use the relative frequency as an estimate for a probability based on that normal distribution. The first two parts of this question are answered by using the standardised normal using the lower and upper sections of the frequency distribution to give relative frequencies in place of probabilities. Using the inverse of Φ gives the values of z in the standardisation equation. Solving these simultaneously gives σ and μ. There is no need to divide the variance by n in this as it is really a relative frequency problem that happens to use a normal distribution. Sample size is irrelevant in relative frequency calculations.
Part iii is saying that the values are estimates as the data is taken from a sample. Thinking back to work done in year 8, you’ll know that a larger sample will give a better estimate of probability when using relative frequencies. A solution for parts i and ii is shown below.
Then we get on to question 7 from the same exam paper. It looks like this:
In this question we again have a sample taken from a normal distribution but don’t have a frequency distribution so can’t use the technique above to find μ and σ. This means that we should calculate those values from the sample summary data given. We need to find the population distribution parameters so will require the unbiased estimators. Using the Population distribution, the probability that a time is over 90 minutes will be the proportion of candidates unable to complete the paper in that time. For part ii of the question the test will be against the sample data so assume the population mean and variance as calculated in part i, but divide the variance by sample size because we’re testing the sample against a population.
In answer to part iii of the question, the assumption of a [normal] distribution is necessary in part i as otherwise we’d not be able to calculate a probability/proportion. It is not necessary in part ii as the Central Limit Theorem applies as the sample size is sufficiently large.

Tags: Teaching, Solutions, posts, School, A Level, Exam, S2
April 03, 2018 at 04:37PM
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