Unit 1 - Number and Algebra
Required Prior Knowledge
Questions
Evaluate:
a) \(2^{2} \)
b) \(2^{3}\)
c) \(2^{4}\)
d) \(2^{5}\)
e) \(2^{6}\)
f) \(3^{2}\)
g) \(3^{3}\)
h) \(3^{4}\)
i) \(3^{5}\)
Solutions
-
a) \(2^{2}=4 \)
b) \(2^{3}=8\)
c) \(2^{4}=16\)
d) \(2^{5}=32\)
e) \(2^{6}=64\)
f) \(3^{2}=9\)
g) \(3^{3}=27\)
h) \(3^{4}=81\)
i) \(3^{5}=243\)
Get Ready
Questions
Solve the equation $$5^{x}=125$$using:
a) knowledge of powers
b) logarithms
Solutions
Notes
When solving equations where the unknown is part of an exponential, there are three main methods we might use:
If the base is the same (or can be made the same easily), then we can equate the powers
If there is a sum of multiples of the same base, we can try a substitution
If the bases are different, then we need to use logarithms
Examples and Your Turns
Example
Solve$$2^{x}=32$$
Your Turn
Solve$$3^{x}=\frac{1}{27}$$
Your Turn
Solve$$2^{x-2}=32$$
Your Turn
Solve$$3^{x+1}=\frac{1}{27}$$
Your Turn
Solve$$4^{x}=32$$
Your Turn
Solve$$9^{x}=\frac{1}{27}$$
Your Turn
Solve$$3^{2-x}=9^{2x}$$
Your Turn
Solve$$5^{3x}=125^{2x-1}$$
Example
Solve$$4^{x}+2^{x}-20=0$$
Your Turn
Solve$$9^{x}-5\times 3^{x}+6=0$$
Example
Solve$$\frac{64^{x}}{2^{x-1}}=\frac{4}{32^{3x}}$$
Your Turn
Solve$$\frac{2^{x-3}}{8^{-x}}=\frac{32}{4^{0.5x}}$$
Example
Solve$$\begin{matrix}3^{x}&\times &9^{y}&=&1\\2^{2x}&\times &4^{y}&=&\frac{1}{8}\end{matrix}$$
Your Turn
Solve$$\begin{matrix}9^{x}&=&27^{y}\\64^{xy}&=&512^{x+1}\end{matrix}$$
Your Turn
Solve$$\frac{16^{x}}{8^{y}}=\frac{1}{4}\\4^{x}2^{y}=16$$
Your Turn
Solve$$4^{a}\sqrt{2^{b}}=64\\\frac{8^{b}}{2^{a}}=4\times\sqrt[3]{2^{a}}$$
Example
Solve$$4^{x}=9$$
Your Turn
Solve$$3^{x-1}=8$$giving your answer in the form \(\frac{\log a}{\log b}\) where \(a,b\) are constants to be found.
-
$$\begin{align}3^{x-1}&=8\\ \log_{3}8&=x-1\\ x&=1+\log_{3}8\end{align}$$
-
$$\begin{align}3^{x-1}&=8\\ \log 3^{x-1} &=\log 8 \\ \left(x-1\right)\log 3 &=\log 8 \\ x-1&=\frac{\log 8}{\log 3} \\ x&=1+\frac{\log 8}{\log 3}\end{align}$$
-
$$\begin{align}3^{x-1}&=8\\ \log 3^{x-1} &=\log 8 \\ \left(x-1\right)\log 3 &=\log 8 \\x\log 3 - \log 3&=\log 8 \\ x\log 3 &=\log 8 + \log 3 \\ x&=\frac{\log 8 + \log 3}{\log 3} \\ x&=\frac{\log 24}{\log 3}\end{align}$$
Your Turn
Solve$$3\times2^{x}=8$$
-
$$\begin{align}3\times 2^{x}&=8\\ 2^{x}&=\frac{8}{3}\\ x&=\log_{2}\frac{8}{3}\end{align}$$
-
$$\begin{align}3\times 2^{x}&=8 \\ 2^{x}&=\frac{8}{3}\\ \log 2^{x} &=\log \frac{8}{3} \\ x\log 2 &=\log \frac{8}{3} \\ x&=\frac{\log 8 - \log 3}{\log 2}\end{align}$$
Your Turn
Solve$$3^{x-4}=24$$
-
$$\begin{align}3^{x-4}&=24\\ x-4&=\log_{3}24\\ x&=4+\log_{3}24\end{align}$$
-
$$\begin{align}3^{x-4}&=24 \\ \log 3^{x-4} &=\log 24 \\ \left(x-4\right)\log 3 &=\log 24 \\ x-4&=\frac{\log 24}{\log 3} \\ x&=4+\frac{\log 24}{\log 3} \end{align}$$
-
$$\begin{align}3^{x-4}&=24 \\ \log 3^{x-4} &=\log 24 \\ \left(x-4\right)\log 3 &=\log 24 \\ x\log 3 -4\log 3&=\log 24 \\ x\log 3&=\log 24 +4\log 3 \\x&=\frac{\log 24 + 4\log 3}{\log 3} \end{align}$$
Your Turn
Solve$$3^{x-1}=2^{x+1}$$
Notes
To solve equations involving logarithms, we usually need to first convert it into an equation in exponential form.
Sometimes we will need to use the Laws of Logarithms first to simplify the logarithm.
We must take care to reject any non-valid solutions, as logarithmic functions are only defined for \(x\gt 0\).
Examples and Your Turns
Example
Solve$$\log_{3}\left(2x-5\right)=2$$
Your Turn
Solve$$\log_{2}\left(3x-5\right)=5$$
-
$$\begin{align}3x-5&=2^{5}\\3x-5&=32\\3x&=37\\x&=\frac{37}{3}\end{align}$$
Your Turn
Solve$$\log_{2}3 +\log_{2}x = 6$$
-
$$\begin{align}\log_{2} 3x&=6\\3x&=2^{6}\\3x&=64\\x&=\frac{64}{3}\end{align}$$
Your Turn
Solve$$\log_{2}5 +\log_{2}y = 4$$
-
$$\begin{align}\log_{2} 5y&=4\\5x&=2^{4}\\5y&=16\\y&=\frac{16}{5}\end{align}$$
Your Turn
Solve$$\log_{3}\left(x+11\right) -\log_{3}\left(x-5\right) = 2$$
-
$$\begin{align}\log_{3} \frac{x+11}{x-5}&=2\\ \frac{x+11}{x-5}&=3^{2}\\ \frac{x+11}{x-5}&=9\\ x+11&=9x-45\\ 56&=8x\\x&=7\end{align}$$
Your Turn
Solve$$\log_{3}\left(x+8\right) -\log_{3}\left(x-3\right) = 3$$
-
$$\begin{align}\log_{3} \frac{x+8}{x-3}&=3\\ \frac{x+8}{x-3}&=3^{3}\\ \frac{x+8}{x-3}&=27\\ x+8&=27x-81\\ 89&=26x\\x&=\frac{89}{26}\end{align}$$
Example
Solve$$\log_{2}x +\log_{2}\left(10-x\right) = 4$$
Your Turn
Solve$$\ln\left(x-2\right) +\ln\left(2x-3\right) = 2\ln x$$
Your Turn
Solve$$2\log_{9}\left(x+1\right)=\log_{9}\left(2x-3\right) +1$$
-
$$\begin{align}\log_{9} \left(x+1\right)^{2}&=\log_{9} \left(2x-3\right) + \log_{9} 9 \\ \log_{9} \left(x+1\right)^{2}&=\log_{9} 9\left(2x-3\right) \\ \left(x+1\right)^{2}&= 9\left(2x-3\right) \\ x^{2}+2x+1&=18x-27 \\ x^{2}-16x+28&=0 \\ \left(x-2\right)\left(x-14\right)&=0\end{align}\\x=2\text{ or }x=14$$
Your Turn
Solve$$2\log_{3}\left(x+2\right)=\log_{3}\left(2x+1\right) +1$$
-
$$\begin{align}\log_{3}\left(x+2\right)^{2}&=\log_{3}\left(2x+1\right) +\log_{3} 3 \\ \log_{3}\left(x+2\right)^{2}&=\log_{3}3\left(2x+1\right) \\ \left(x+2\right)^{2}&=3\left(2x+1\right) \\ x^{2}+4x+4&&=6x+3\\x^{2}-2x+1&=0 \\ \left(x-1\right)^{2}&=0 \end{align}\\x=1$$
Your Turn
Solve$$\log_{6}x +\log_{6}\left(x-5\right) =1$$
-
$$\begin{align}\log_{6}x\left(x-5\right)&=\log_{6}6 \\ x\left(x-5\right)&=6 \\ x^{2}-5x-6&=0 \\ \left(x-6\right)\left(x+1\right)&=0\end{align}\\ \therefore x=6\text{ or }x=-1$$But \(x=-1\) is not valid in the original equation as we cannot evaluate \(\log_{6} -1\). Thus$$x=6$$
Your Turn
Solve$$\log_{5}\left(5x\right) -\log_{5}\left(x+2\right)=\log_{5}\left(x+6\right) -\log_{5}x$$
-
$$\begin{align}\log_{5}\frac{5x}{x+2}&=\log_{5}\frac{x+6}{x}\\ \frac{5x}{x+2}&=\frac{x+6}{x}\\ 5x^{2}&=\left(x+6\right)\left(x+2\right) \\ 5x^{2}&=x^{2}+8x+12 \\4x^{2}-8x-12&=0 \\x^{2}-2x-3&=0 \\ \left(x-1\right)\left(x-2\right)&=0 \end{align}\\ \therefore x=1\text{ or }x=2$$
Your Turn
Solve$$\log_{3}x -4\log_{x}3 +3=0$$
Your Turn
Solve$$\begin{matrix}\log_{3}x &+&4\log_{9}y &=&2\\ 2\log_{4}x &+&\log_{2}y &=&1\end{matrix}$$
Key Facts
Use this applet to generate a prompt for a Key Fact that you need to know for the course. The idea is that you should KNOW these key facts in order to be able to solve problems.