Exponentials and logarithms – Year 12 A level maths revision notes to download as PDF

Worked example 1

Write \(2\log3+\log4\) as a single logarithm.

\(\log9+\log4=\log36\).

Worked example 2

Solve \(3^x=20\) and \(3^x=2^{x+1}\).

\(3^x=20\Rightarrow x=\dfrac{\ln20}{\ln3}=2.73\). For the second: \(x\ln3=(x+1)\ln2\), \(x(\ln3-\ln2)=\ln2\), so \(x=\dfrac{\ln2}{\ln1.5}=1.71\).

Worked example 3

Solve \(\log_2(x+3)+\log_2(x-3)=4\).

\(\log_2(x^2-9)=4\Rightarrow x^2-9=16\Rightarrow x=\pm5\). Rejecting \(x=-5\) (negative arguments) leaves \(x=5\).

Worked example 4

Solve \(2^{2x}-6\times2^x+8=0\).

Let \(y=2^x\). Then \(y^2-6y+8=0\), so \(y=2\) or \(4\). Hence \(x=1\) or \(x=2\).

Worked example 5

A graph of \(\log_{10}y\) against \(x\) passes through \((0,1)\) and \((2,1.954)\). Find \(y\) in terms of \(x\).

Intercept \(1=\log a\Rightarrow a=10\). Gradient \(\dfrac{0.954}{2}=0.477=\log b\Rightarrow b=3\). So \(y=10\times3^x\).

Worked example 6

\(y=ax^n\) gives \(y=12\) at \(x=2\) and \(y=192\) at \(x=8\). Find \(a\) and \(n\).

Divide: \(16=4^n\), so \(n=2\), then \(a=3\). A log-log graph would show a line of gradient 2.

Worked example 7

\(P=200e^{0.03t}\). Find \(P(10)\) and the doubling time.

\(P(10)=200e^{0.3}\approx270\), and the population doubles after \(\dfrac{\ln2}{0.03}=23.1\) years.

Worked example 8

£2000 earns 3% per year. When does it first exceed £3000?

Solve \(1.03^n>1.5\): \(n>\dfrac{\ln1.5}{\ln1.03}=13.7\), so 14 full years. Dividing by \(\ln1.03\) is safe because it is positive; dividing by a negative log would reverse the inequality.

Worked example 9

The mass of a sample is \(N=800e^{-0.04t}\) (grams, \(t\) in days). Find the half-life and the mass after 20 days.

\(e^{-0.04t}=\tfrac12\), so \(t=\dfrac{\ln2}{0.04}=17.3\) days. After 20 days \(800e^{-0.8}\approx359\) g.

Worked example 10: Using the laws to simplify

Given that \(\log_2a=p\) and \(\log_2b=q\), write \(\log_2\dfrac{8a^2}{\sqrt b}\) in terms of \(p\) and \(q\).

\(\log_2\dfrac{8a^2}{\sqrt b}=\log_28+\log_2a^2-\log_2b^{1/2}=3+2\log_2a-\tfrac12\log_2b=3+2p-\tfrac12q\). For a numerical check take \(a=4\) and \(b=16\), so \(p=2\) and \(q=4\): the expression \(\dfrac{8\times16}{4}=32\) has \(\log_232=5\), and \(3+2\times2-\tfrac12\times4=5\) ✓. Each law of logarithms corresponds to a law of indices: a product becomes a sum, a quotient a difference and a power a multiple.

Worked example 11: Changing the base

Use the change of base formula to evaluate \(\log_725\) to 3 significant figures and to show that \(\log_82=\tfrac13\).

\(\log_ab=\dfrac{\log_cb}{\log_ca}\) for any base \(c\). So \(\log_725=\dfrac{\ln25}{\ln7}=\dfrac{3.2189}{1.9459}=1.65\). Also \(\log_82=\dfrac{\log_22}{\log_28}=\dfrac13\). This is why scientific calculators only need \(\log_{10}\) and \(\ln\) keys. Check: \(7^{1.654}=25.0\).

Worked example 12: A model fitted by taking logarithms

The value of a collectable stamp was £12 in 2010 and £34 in 2020. Assuming exponential growth \(V=ab^t\) (\(t\) years from 2010), find \(a\) and \(b\) and predict the value in 2030.

\(a=12\) since \(V(0)=12\). \(V(10)=34=12b^{10}\), so \(b^{10}=\tfrac{34}{12}=2.8333\) and \(b=2.8333^{0.1}=1.1098\): the annual growth is about 11%. In 2030 \(t=20\) and \(V=12\times1.1098^{20}=12\times8.027=£96.3\). Equivalently \(V=12\times\left(\tfrac{34}{12}\right)^2=12\times8.03=96.3\). Taking logs, \(\log V=\log12+t\log b\) is a straight line with gradient \(0.1\times\log\tfrac{34}{12}=0.0452\) per year.

Worked example 13: A trap in solving logarithmic equations

Solve \(\log_3(x-4)+\log_3(x+4)=2\) and explain which solution must be rejected.

\(\log_3\left[(x-4)(x+4)\right]=2\) gives \(x^2-16=9\), so \(x^2=25\) and \(x=\pm5\). For \(x=-5\) the terms \(x-4=-9\) and \(x+4=-1\) are negative, and the logarithm of a negative number is undefined, so \(x=-5\) is rejected. The only solution is \(x=5\). Check: \(\log_31+\log_39=0+2=2\) ✓. Always check every solution in the original equation, because combining logarithms can introduce solutions that do not belong.

Worked example 14: Exponential equations that need a substitution and logs

Solve \(e^{2x}-4e^x-5=0\) and explain why one root is rejected.

Let \(y=e^x\): \(y^2-4y-5=0\), \((y-5)(y+1)=0\). \(e^x=5\) gives \(x=\ln5=1.61\). \(e^x=-1\) has no solution, because \(e^x\) is always positive. So the only solution is \(x=\ln5\). Check: \(e^{2\ln5}-4\times5-5=25-20-5=0\) ✓. The graph of \(e^{2x}-4e^x-5\) crosses the \(x\)-axis only once, at \(x\approx1.61\).

Worked example 15: Reading growth and decay from an equation

Compare \(A=300e^{0.04t}\) and \(B=500e^{-0.03t}\) and find when \(A=B\).

\(A\) is growing at a rate of about 4% per unit time and \(B\) is decaying at about 3%. They are equal when \(300e^{0.04t}=500e^{-0.03t}\), so \(e^{0.07t}=\tfrac53\) and \(t=\dfrac{\ln(5/3)}{0.07}=\dfrac{0.5108}{0.07}=7.30\). At that time both equal \(300e^{0.292}=401.9\). Since \(A\) grows and \(B\) falls, \(A\) stays larger afterwards.

Worked example 16: Percentage growth rates and the number e

A sum of money grows by 5% per year. Write the model in the form \(V=V_0e^{kt}\) and find the time to triple.

Growth of 5% a year gives \(V=V_0\times1.05^t\). To write this with base \(e\) note that \(1.05=e^{\ln1.05}\), so \(V=V_0e^{0.04879t}\) and \(k=\ln1.05=0.0488\). Tripling needs \(e^{kt}=3\), so \(t=\dfrac{\ln3}{0.04879}=22.5\) years. Using \(1.05^t=3\) and \(t=\dfrac{\log3}{\log1.05}\) gives exactly the same answer, which shows that the two forms of the model are equivalent. A continuous growth rate \(k\) is always slightly smaller than the matching annual percentage rate, because compounding happens all the time.

Worked example 17: Finding the intersection of an exponential and a straight line numerically

Show that the equation \(2^x=x+3\) has a root between 2 and 3 and use a search to find it to 1 decimal place.

Let \(f(x)=2^x-x-3\). Then \(f(2)=4-2-3=-1<0\) and \(f(3)=8-3-3=2>0\), so the sign changes and there is a root in between. Try \(f(2.5)=5.657-5.5=0.157>0\) and \(f(2.4)=5.278-5.4=-0.122<0\), so the root is between 2.4 and 2.5. \(f(2.45)=5.464-5.45=0.014>0\) so the root is below 2.45, meaning it is 2.4 to 1 decimal place (about 2.446). Equations mixing exponentials and linear terms usually cannot be solved with logarithms alone, so a numerical search or a graph is needed.