John Napier (1550–1617) was a Scottish mathematician, physicist and astronomer, best known for his discovery of logarithms. Napier also popularised the use of the decimal point and invented what came to be known as Napier's bones, a manually-operated device that could calculate products and quotients. The discovery of logarithms made possible the later development of the slide rule, an essential engineers' calculating tool until the 1970s, when electronic calculators took over. Napier's birthplace, Merchiston Tower in Edinburgh, is now owned by Edinburgh Napier University, which was named in his honour.
From the question, \(x\gt 0\) so that we aren't trying to take the logarithm of a negative number. So we should disregard \(x=-9\). The only solution is \(x=4.\)
Example 6 (non-calculator)
Given that \(log_a\,45-log_a\,5=\large\frac{1}{2}\normalsize,\) find the value of \(a.\)
In this type of logarithmic equation, we need to find the base. It's much easier than the previous example.
Solve \(6^{2x+1}=19,\) correct to 3 significant figures.
This is an example of an exponential equation because the unknown is within the power.
The method is to take logs of both sides. Any base will work, but your calculator can do base \(10\) \((log)\) or base \(e\) (\(ln)\) so obviously it's best to use one of those!
The number of people infected by a virus is described by the exponential formula \(I(t)=5e^{0.07t},\) where \(t\) is the time in days. (a) How many people are infected on day zero? (b) To the nearest day, how long will it take for the number of infections to double?
(a) \(I(0)=5e^{0.07\times 0}=5e^0=5\) people.
(a) \(2\times 5=10\) so we substitute \(I(t)=10\) and solve for \(t.\)
Note that the base is \(e\) so we will be using natural logarithms: \(log_e\) or \(ln.\)
Now we look at the graph and think about the equation of the straight line. The general equation of a straight line is \(y=mx+c\) but in this example the vertical axis is labelled \(log_5\,y\), not \(y.\) So it's actually \(log_5\,y=mx+c.\)
We can obtain the gradient \(m\) from the two points given. \(m = \large\frac{1-(-1)}{4-0}\normalsize = \large\frac{1}{2}\normalsize.\)
From the graph, the \(y\)-intercept \(c=-1.\)
So the equation is \(log_5\,y=\large\frac{1}{2}\normalsize x-1.\:\:②\)
To complete this, we compare equations ① and ② term by term.
\(log_5\,b=\large\frac{1}{2}\normalsize\) so \(b=5^{1/2}=\sqrt{5}.\)
\(log_5\,a=-1\) so \(a=5^{-1}=\large\frac{1}{5}.\)
Example 10 (non-calculator)
Two variables, \(x\) and \(y\), are connected by the equation \(y=kx^{n}.\) The graph of \(log_3\,y\) against \(log_3\,x\) is shown below.
Find the values of \(k\) and \(n.\)
We start by expanding the equation \(y=kx^n\) as follows:
Now we look at the graph and think about the equation of the straight line. The general equation of a straight line is \(y=mx+c\) but in this example the axes are labelled \(log_3\,y\) and \(log_3\,x\), not \(y\) and \(x.\) So it's actually \(log_3\,y=m\,log_3\,x+c.\)
We can obtain the gradient \(m\) from the two points given. \(m=\large\frac{3}{4}\normalsize.\)
Also from the graph, the \(y\)-intercept \(c=3.\)
So the equation is \(log_3\,y=\large\frac{3}{4}\normalsize log_3\,x+3.\:\:②\)
To complete this, we compare equations ① and ② term by term.
Comparing the \(log_3\,x\) terms, we see that \(n=\large\frac{3}{4}\normalsize.\)
Comparing the constant terms, \(log_3\,k=3\) so \(k=3^3=27.\)
Example 11 (non-calculator)
SQA Higher Maths 2017 Paper 1 Q12
Given that \(log_a\,36-log_a\,4=\large\frac{1}{2}\normalsize,\) find the value of \(a.\)
We need to find the base, so we apply the log laws as follows: