Higher Maths
Logarithms and Exponentials

This introductory example is only checking your basic understanding of what a logarithm is.

Definition: \(\text{log}_b\,x\) means the power to which we must raise the base \(b\) to get \(x\small.\)

\(4^3=64\) so \(\text{log}_4\,64=3\small.\)

This would be worth three marks. It tests your understanding of what logarithm is, as well as two log laws:

• \(\text{log}_b\,a^n = n\,\text{log}_b\,a\)
• \(\text{log}_b\,x+\text{log}_b\,y = \text{log}_b\,xy\)

Note also that \(\text{log}\,x\) written without any base means \(\text{log}_{10}\,x\small.\)

So, putting all this together, here is the solution:

$$ \begin{eqnarray} \text{log}\,40+2\,\text{log}\,5 &=& \text{log}\,40+\text{log}\,5^2 \\[6pt] &=& \text{log}\,40+\text{log}\,25 \\[6pt] &=& \text{log}\,(40\times 25) \\[6pt] &=& \text{log}\,1000 \\[6pt] &=& 3 \end{eqnarray} $$

Note: The last step is because \(10^3 = 1000\small.\)

Like the previous example, this would be worth three marks. It tests your understanding of the definition of logarithm, as well as two log laws:

• \(\text{log}_b\,a^n = n\,\text{log}_b\,a\)
• \(\text{log}_b\,x-\text{log}_b\,y = \text{log}_b\,\frac{x}{y}\)

So the solution looks something like this:

$$ \begin{eqnarray} \text{log}_9\,12-2\,\text{log}_9\,2 &=& \text{log}_9\,12-\text{log}_9\,2^2 \\[6pt] &=& \text{log}_9\,12-\text{log}_9\,4 \\[6pt] &=& \text{log}_9\,\small\frac{12}{4}\normalsize \\[6pt] &=& \text{log}_9\,3 \\[6pt] &=& \small\frac{1}{2}\normalsize \\[6pt] \end{eqnarray} $$

Note: The last step is because \(9^{1/2} = \sqrt{9} = 3\small.\)

This is another three-mark question using the log laws mentioned above.

$$ \begin{align*} 2\,\text{log}_6\,3 + \smash{\tfrac{1}{3}}\,\text{log}_6\,64 &= \text{log}_6\,3^2 + \smash{\text{log}_6\,64^{1/3}} \\[10pt] &= \text{log}_6\,9 + \smash{\text{log}_6\,\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{64}} \\[10pt] &= \text{log}_6\,9 + \text{log}_6\,4 \\[10pt] &= \text{log}_6\,(9 \times 4) \\[10pt] &= \text{log}_6\,36 \\[10pt] &= 2 \end{align*} $$

The strategy here is to apply log laws until we can turn it into a purely algebraic equation, without any logarithms.

$$ \begin{eqnarray} \text{log}_6\,x+\text{log}_6\,(x+5) &=& 2 \\[6pt] \text{log}_6\,x(x+5) &=& 2 \\[6pt] \end{eqnarray} $$

From here, there are two equally good ways to continue the solution.

Method 1: Write the \(2\) as \(\text{log}_6\,36\) so that you can remove the logs in the next line:

$$ \begin{eqnarray} \text{log}_6\,x(x+5) &=& \text{log}_6\,36 \\[6pt] x(x+5) &=& 36 \\[12pt] \end{eqnarray} $$

Method 2: Use the definition of logarithm to change the equation into exponential form:

$$ \begin{gather} \text{log}_6\,x(x+5)=2 \\[6pt] 6^2=x(x+5) \\[6pt] 36=x(x+5) \\[12pt] \end{gather} $$

Whichever of these you prefer, we now have a quadratic equation. So we complete the solution like this:

$$ \begin{gather} x^2+5x =36 \\[6pt] x^2+5x-36=0 \\[6pt] (x-4)(x+9)=0 \\[6pt] x-4=0\:\:\small\textsf{or}\normalsize\:\:x+9=0 \\[6pt] x=4\:\:\small\textsf{or}\normalsize\:\:x=-9 \\[6pt] \end{gather} $$

From the question, \(x\gt 0\) so that we aren't trying to take the logarithm of a negative number. So we should disregard \(x=-9\small.\) The only solution is \(x=4\small.\)

In this type of logarithmic equation, we need to find the base. It's much easier than the previous example.

$$ \begin{gather} \text{log}_a\,45-\text{log}_a\,5=\small\frac12\\[9pt] \text{log}_a\,\small\frac{45}{5}\normalsize=\small\frac12\\[9pt] \text{log}_a\,9=\small\frac12 \\[9pt] a^{1/2}=9\\[9pt] \sqrt{a}=9\\[9pt] a=9^2\\[9pt] a=81 \end{gather} $$

Note: The exam board has previously designed questions of this type to feature a square number, such as the \(9\) above. This is to give weaker students the opportunity to fall into a trap, by square rooting the \(9\) instead of squaring it, thereby giving a wrong answer of \(3\small.\) Take care!

The unknown is within the exponent, so this is an example of an exponential equation.

The method is to take logs of both sides. Any base will work. We will use base \(e\small.\)

$$ \begin{gather*} 6^{2x+1}=19\\[12pt] \text{ln}\,6^{2x+1}=\text{ln}\,19\\[12pt] (2x+1)\,\text{ln}\,6=\text{ln}\,19\\[12pt] 2x+1=\smash{\small\frac{\text{ln}\,19}{\text{ln}\,6}\normalsize}\\[15pt] 2x=\smash{\small\frac{\text{ln}\,19}{\text{ln}\,6}\normalsize-1}\\[15pt] x=\smash{\left(\small\frac{\text{ln}\,19}{\text{ln}\,6}\normalsize-1\right)\div 2}\\[15pt] x \approx 0.322 \end{gather*} $$

(a)  \(I(0)=5e^{0.07\times 0}=5e^0=5\) people.

(b)  \(2\times 5=10\) so we substitute \(I(t)=10\) and solve for \(t.\)

Note that the base is \(e\) so we will be using natural logarithms: \(\text{log}_e\small,\) also known as \(\text{ln}\small.\)

$$ \begin{eqnarray} I(t) &=& 5e^{0.07t} \\[6pt] 10 &=& 5e^{0.07t} \\[6pt] 2 &=& e^{0.07t} \\[6pt] \text{ln}\,2 &=& \text{ln}\ e^{0.07t} \\[6pt] \text{ln}\,2 &=& 0.07t \\[6pt] t &=& \small\frac{\text{ln}\,2}{0.07}\normalsize \\[6pt] t &\approx& 9.902 \\[6pt] \end{eqnarray} $$

So, to the nearest day, it will take \(10\) days for the number of infections to double.

We start by expanding the equation \(y=ab^x\) as follows:

$$ \begin{eqnarray} y &=& ab^x \\[6pt] \text{log}_5\,y &=& \text{log}_5\,ab^x \\[6pt] \text{log}_5\,y &=& \text{log}_5\,a+\text{log}_5\,b^x \\[6pt] \text{log}_5\,y &=& \text{log}_5\,a+x\,\text{log}_5\,b\:\:① \end{eqnarray} $$

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 \(\text{log}_5\,y\), not \(y\small.\) So it's actually \(\text{log}_5\,y=mx+c\small.\)

We can obtain the gradient \(m\) from the two points given: \(m = \large\frac{1-(-1)}{4-0}\normalsize = \large\frac{1}{2}\small.\)

From the graph, the \(y\)-intercept \(c=-1\small.\)

So the equation is \(\text{log}_5\,y=\large\frac{1}{2}\normalsize x-1\small.\normalsize\:\:②\)

To complete this, we compare equations \(①\) and \(②\) term-by-term.

\(\text{log}_5\,b=\large\frac{1}{2}\normalsize\) so \(b=5^{1/2}=\sqrt{5}\,\small.\)

\(\text{log}_5\,a=-1\) so \(a=5^{-1}=\large\frac{1}{5}\small.\)

We start by expanding the equation \(y=kx^n\) as follows:

$$ \begin{eqnarray} y &=& kx^n \\[6pt] \text{log}_3\,y &=& \text{log}_3\,kx^n \\[6pt] \text{log}_3\,y &=& \text{log}_3\,k+\text{log}_3\,x^n \\[6pt] \text{log}_3\,y &=& \text{log}_3\,k+n\,\text{log}_3\,x\:\:① \end{eqnarray} $$

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 \(\text{log}_3\,y\) and \(\text{log}_3\,x\), not \(y\) and \(x\small.\) So it's actually \(\text{log}_3\,y=m\,\text{log}_3\,x+c\small.\)

We can obtain the gradient \(m\) from the two points given. \(m=\large\frac{3}{4}\small.\)

Also from the graph, the \(y\)-intercept \(c=3\small.\)

So the equation is \(\text{log}_3\,y=\large\frac{3}{4}\normalsize \text{log}_3\,x+3\small.\normalsize\:\:②\)

To complete this, we compare equations \(①\) and \(②\) term-by-term.

Comparing the \(\text{log}_3\,x\) terms, we see that \(n=\large\frac{3}{4}\small.\)

Comparing the constant terms, \(\text{log}_3\,k=3\) so \(k=3^3=27\small.\)

We need to find the base, so we apply the log laws as follows:

$$ \begin{gather} \text{log}_a\,36-\text{log}_a\,4=\small\frac{1}{2}\normalsize\\[6pt] \text{log}_a\,\small\frac{36}{4}\normalsize=\small\frac{1}{2}\normalsize\\[6pt] \text{log}_a\,9=\small\frac{1}{2}\normalsize\\[6pt] a^{1/2}=9\\[6pt] \sqrt{a}=9\\[6pt] a = 81 \end{gather} $$

As usual with these question types, we apply the log laws to combine the terms:

$$ \begin{align*} \text{log}_5\,250-\smash{\small\frac{1}{3}\normalsize}\,\text{log}_5\,8 &= \text{log}_5\,250-\text{log}_5\,8^{1/3} \\[9pt] &= \smash{\text{log}_5\,250-\text{log}_5\,\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{8}} \\[9pt] &= \text{log}_5\,250-\text{log}_5\,2 \\[9pt] &= \text{log}_5\,\smash{\small\frac{250}{2}\normalsize} \\[9pt] &= \text{log}_5\,125 \\[9pt] &= 3 \end{align*} $$

This 4-mark question tested understanding of the definition of logarithm, as well as three log laws:

• \(\text{log}_b\,a^n = n\,\text{log}_b\,a\)
• \(\text{log}_b\,x+\text{log}_b\,y = \text{log}_b\,xy\)
• \(\text{log}_b\,x-\text{log}_b\,y = \text{log}_b\,\frac{x}{y}\)

So the solution looks like this:

$$ \begin{eqnarray} &\phantom{=}& \!\!\!\!\!\!\text{log}_2\,6+\,\text{log}_2\,12-2\,\text{log}_2\,3 \\[6pt] &=& \text{log}_2\,6+\,\text{log}_2\,12-\text{log}_2\,3^2 \\[6pt] &=& \text{log}_2\,6+\,\text{log}_2\,12-\text{log}_2\,9 \\[6pt] &=& \text{log}_2\,\small\frac{6\times 12}{9}\normalsize \\[6pt] &=& \text{log}_2\,8 \\[6pt] &=& 3 \end{eqnarray} $$

This question tested understanding of the definition of logarithm, as well as two log laws:

• \(\text{log}_b\,a^n = n\,\text{log}_b\,a\)
• \(\text{log}_b\,x-\text{log}_b\,y = \text{log}_b\,\frac{x}{y}\)

So the solution looks like this:

$$ \begin{eqnarray} 2\,\text{log}_3\,6-\,\text{log}_3\,4 &=& \text{log}_3\,6^2-\text{log}_3\,4 \\[6pt] &=& \text{log}_3\,36-\text{log}_3\,4 \\[6pt] &=& \text{log}_3\,\small\frac{36}{4}\normalsize \\[6pt] &=& \text{log}_3\,9 \\[6pt] &=& 2 \end{eqnarray} $$

First we use the relevant log law to reduce the left hand side to a single term:

$$ \begin{eqnarray} \text{log}_5\,x-\text{log}_5\,3 &=& 2 \\[6pt] \text{log}_5\,\small\frac{x}{3}\normalsize &=& 2 \\[6pt] \end{eqnarray} $$

Now, we can use the definition of logarithm to write this in exponential form:

$$ \small\frac{x}{3}\normalsize =5^2 $$ $$ \small\frac{x}{3}\normalsize =25 $$ $$ x = 25\times 3 $$ $$ x = 75 $$

Here, we need to use the log laws to simplify the expression into a single term:

$$ \begin{eqnarray} &\:& \!\!\!\!\!\!\text{log}_a\,5+\text{log}_a\,80-2\,\text{log}_a\,10\\[10pt] &=& \text{log}_a\,5+\text{log}_a\,80-\text{log}_a\,{10^2}\\[10pt] &=& \text{log}_a\,5+\text{log}_a\,80-\text{log}_a\,100\\[10pt] &=& \text{log}_a\,{\small\frac{5\times 80}{100}}\\[10pt] &=& \text{log}_a\,{\small\frac{400}{100}}\\[10pt] &=& \text{log}_a\,4 \\[10pt] \end{eqnarray} $$

(a)  At the end of 2020, \(t=0\small,\) so:

$$ \begin{eqnarray} N &=& 6.8e^{kt} \\[6pt] &=& 6.8e^{0} \\[6pt] &=& 6.8 \\[6pt] \end{eqnarray} $$

So the estimated number of electric vehicles worldwide at the end of 2020 was \(6.8\) million.

(b)  At the end of 2030, \(t=10\) and \(N=125\small,\) so:

$$ \begin{eqnarray} N &=& 6.8e^{kt} \\[6pt] 125 &=& 6.8e^{10k} \\[6pt] \small\frac{125}{6.8}\normalsize &=& e^{10t} \\[6pt] \text{ln}\,\left(\!\small\frac{125}{6.8}\normalsize\!\right) &=& \text{ln}\,\left(e^{10k}\right) \\[6pt] \text{ln}\,\left(\!\small\frac{125}{6.8}\normalsize\!\right) &=& 10k \\[6pt] k &=& \small\frac{1}{10}\,\text{ln}\,\left(\!\small\frac{125}{6.8}\normalsize\!\right) \\[6pt] k &\approx& 0.29114 \\[6pt] \end{eqnarray} $$

This 3-mark question requires a firm understanding of the N5 index laws, the definition of logarithm and some of the laws of logarithms.

The solution may be presented as follows:

$$ \begin{eqnarray} 3\,\text{log}_3\,2+\text{log}_3\,\small\frac{1}{24}\normalsize &=& \text{log}_3\,2^3+\text{log}_3\,\small\frac{1}{24}\normalsize \\[8pt] &=& \text{log}_3\,8+\text{log}_3\,\small\frac{1}{24}\normalsize \\[8pt] &=& \text{log}_3\left(8\times\small\frac{1}{24}\normalsize\right) \\[8pt] &=& \text{log}_3\left(\small\frac{1}{3}\normalsize\right) \\[8pt] &=& \text{log}_3\,\left(3^{-1}\right) \\[8pt] &=& -\!1 \\[8pt] \end{eqnarray} $$

We need to find the base, so we rearrange and apply the log laws as follows:

$$ \begin{gather} \text{log}_a\,75=2+\text{log}_a\,3\\[8pt] \text{log}_a\,75-\text{log}_a\,3=2\\[8pt] \text{log}_a\,\small\frac{75}{3}\normalsize=2\\[8pt] \text{log}_a\,25=2\\[8pt] a^2=25 \\[8pt] a=5 \end{gather} $$

Note: Although \(-5\) is also a solution of \(a^2=25\small,\) we should ignore this as we were told in the question that \(a \gt 0\small.\) Moreover, within the real number set, a logarithm can never have a negative base.

(a)  The initial mass is when \(t=0\small,\) so we substitute:

$$ \begin{eqnarray} M &=& 150\,e^{\large{-0.0054t}} \\[6pt] &=& 150\,e^{0} \\[6pt] &=& 150 \\[6pt] \end{eqnarray} $$

So the initial mass of the radioactive substance is \(150\) micrograms.

(b)  We substitute \(M=120\) and solve the exponential equation:

$$ \begin{eqnarray} M &=& 150\,e^{\large{-0.0054t}} \\[8pt] 120 &=& 150\,e^{\large{-0.0054t}} \\[8pt] \small\frac{120}{150}\normalsize &=& e^{\large{-0.0054t}} \\[8pt] 0.8 &=& e^{\large{-0.0054t}} \\[8pt] \text{ln}\,0.8 &=& \text{ln}\,e^{\large{-0.0054t}} \\[8pt] \text{ln}\,0.8 &=& -0.0054t \\[8pt] t &=& \small\frac{\text{ln}\,0.8}{-0.0054} \\[8pt] t &\approx& 41.322\cdots \\[8pt] \end{eqnarray} $$

So it takes approximately \(41.3\) years for the radioactive substance to decay to \(120\) micrograms.