Logarithms
Logarithms (logs, informally), are the opposite of exponential functions - this is what everybody told me when I learned them. But I couldn't ever make sense of that. What on earth is the 'opposite' of a mathmatical expression? It makes no sense.
What I think these teachers mean, is that a log is the inverse of it's exponential function.
Here is the graph of \(y = 2^x \ \text{(red),} \ \ y = x \ \text{(dashed green),} \ \ y = \log_2{x} \ \text{(red)} \)
Where:
\(a^x = y\),
\(\log_{a}{y} = x \text{, given } y > 0\)
This log simply means that x is equaly to the number of times a is multiplied by itself to equal y. This is all a log is.
Confidence of this concept comes with practise, and remember that some log laws are on the formula sheet.
In the example above, the position of \(a\) is reffered as the 'base' of the log. If no base is present, the log is named a 'common log', and has a true base of 10 - you just can't see it.
Log Scales
A log scale is one in which increment of the graph increases exponentially.
This is a 1D number line in linear scale:
This is a 2D graph in linear scale:
This is a 1D number line in logarithmic scale:
This is a 2D graph in logarithmic scale:
Log Laws
1. \(\log_a{a} =1 \)
2. \(\log_a{1} =0 \)
3. \(\log_a{(b^n)} =n \log_a{b} \)
4. \(\log_a{(bc)} = \log_a{b} + \log_a{c} \)
5. \(\log_a{(\frac{b}{c})} = \log_a{b} - \log_a{c} \)
6. \(\log_a{(\frac{1}{b})} = -\log_a{b} \)
1) Rewrite in exponential form:
a. \(log_{25}{625} =2\)
\({25}^2 = 625\)
b. \(log_2{256} =8\)
\(2^8 = 256\)
c. \(log_{10}{1000} =3\)
\(10^3 = 1000\)
2) Rewrite in logarithmic form:
a. \(7^3 = 343\)
\(\log_7{343} = 3 \)
b. \(\frac{1}{16} = 2^{-4}\)
\(\begin{aligned} \log_2{(\frac{1}{16} )} &= -4 \\[5pt] -\log_2{16} &= -4 \\[5pt] \log_2{16} &= 4 \end{aligned} \)
c. \(\displaystyle 9 = \sqrt{81} \)
\(\log_{81}{9} = 0.5 \)
3) Solve
a. \(\log_{3}{(\frac{3}{81})} \)
\(\begin{aligned} &= \log_3{3} - \log_3{3^4} \\[5pt] &= 1 - 4 \\[5pt] &= -3 \end{aligned} \)
Logarithmic Functions
A variable may be included in an expression involving logs.
These functions are very easy and with few tricks: simply just expand the logs using log laws, until you're able to convert to exponential form, then solve the expression in terms of the variable required.
4)
a. \(\log{x^2+36} = 2 \)
\(\begin{aligned} 10^2 &= x^2 + 36 \\ x^2 &= 100 - 36 \\ &= 64 \\ x &= \pm 8 \end{aligned} \)
b. \(\log_2{x-3} = 6\)
\(\begin{aligned} 2^6 &= x-3 \\ x &= 2^6 + 3 \\ &= 64 + 3 \\ &= 67 \end{aligned} \)
c. \(\frac{1}{2} \log{x} + \log{(1+x^2)} - 0.25\log{(x^2)} = \log{(37)} \)
\(\displaystyle \begin{aligned} \log{(37)} &= \log{x^{0.5}} + \log{(1+x^2)} - \log{ \left[(x^2)^{0.25} \right]} \\ &= \log{(x^{0.5} + x^{2.5})} - \log{ \left[(x^2)^{0.25} \right]} \\ &= \log{ \left( \cfrac{x^{0.5} + x^{2.5}}{x^{0.5}} \right) } \\ \log{(x^0 + x^2)} &= \log{(36)} \\ 1+ x^2 &= 37 \\ x^2 &= 36 \\ x &= \pm 6 \\ x &= 6 \text{, as } x \text{ can't be negative}\end{aligned} \)
Logs may be used to solve equations involving exponents of different bases, simply just take the log of both sides of the equation using an appropriate base:
5) Solve for \(x\) in exact form
a. \(3^x = 27\)
\(\begin{aligned} \log_3{3^x} &= \log_3{27} \\[5pt] x\log_3{3} &= 3 \\[5pt] x &= 3 \end{aligned} \)
b. \(4^{2x+3} = 7 \text{ using the common logarithm}\)
\(\begin{aligned} \log_{10}{4^{2x+3}} &= \log_{10}{7} \\[5pt] (2x+3)\log_{}{4} &= \log_{}{7} \\[5pt] 2x\log{4} + \log{(4^3)} &= \log{7} \\[5pt] 2x\log{4} &= \log{7} - \log{64} \\[5pt] 2x\log{4} &= \log{7} - 2\log{8} \\[5pt] x &= \frac{\log{7} - 2\log{8}}{2\log{4}} \\[5pt] &= \frac{\log{7}}{2\log{4}} - \frac{2\log{8}}{2\log{4}} \\[5pt] &= \frac{\log{7}}{4\log{2}} - \frac{2\log{(2^3)}}{2\log{(2^2)}} \\[5pt] &= \frac{\log{7}}{4\log{2}} - \frac{6\log{2}}{4\log{2}} \\[5pt] &= \frac{\log{7}}{4\log{2}} - \frac{3}{2} \end{aligned} \)
c. \(10(7^{2x}) = 50(7^{x-1})\)
\(\begin{aligned} 7^{2x} &= 5(7^{x-1}) \\[5pt] \frac{7^{2x}}{7^{x-1}} &= 5 \\[5pt] 7^{2x-(x-1)} &= 5 \\[5pt] 7^{x+1} &= 5 \\[5pt] \log{7^{x+1}} &= \log{5} \\[5pt] (x+1)\log{7} &= \log{5} \\[5pt] x + 1 &= \frac{\log{5}}{\log{7}} \\[5pt] x &= \frac{\log{5}}{\log{7}} -1 \end{aligned} \)
Applications of Logarithms
Log scales are used in various real-world applications that involve measuring exponential growth, for example:
- pH / number of hydronium ions/hydrogen ions present in solution
- Spread of a virus amongst a population (COVID moment)
- The Richter magnitude scale
- The Decibal Scale
6) The force of an earthquake can be measured using the Richter scale - which is a logarithmic scale that measures relative intensities \(I\), of earthquakes. The scale is defined by \(R = \log{\frac{I}{I_0}}\), where \(I_0\) is the benchmark intensity: the minimum intensity of the fraction.
a. What is the richter scale reading of an earthquake with intensity \(125 I_0\). (calc assumed)
\(\begin{aligned} R &= \log{\frac{125 I_0}{I_0}} \\[5pt] &= \log{125} \\[5pt] &\approx 2.10 \end{aligned} \)
b. What is the intensity of an earthquake that measures 4.5 on the richter scale?
\(\begin{aligned} 4.5 &= \log{(\frac{I}{I_0})} \\[5pt] \frac{I}{I_0} &= 10^{4.5} \\[5pt] I &= 398.11 I_0 \end{aligned} \)
c. Earthquake A, measures 4 on the richter scale. Earthquake B, measures 3.5 on the richter scale. How many times as intense is earthquake A then B?
\(\begin{aligned} 4.5 &= \log{(\frac{I_A}{I_0})} \\[5pt] 10^{4.5} &= \log{(\frac{I_A}{I_0})} \\[5pt] 10^{4.5} I_0 &= I_A \end{aligned} \)
\(\begin{aligned} 3.2 &= \log{(\frac{I_B}{I_0})} \\[5pt] 10^{3.2} &= \log{(\frac{I_B}{I_0})} \\[5pt] 10^{3.2} I_0 &= I_B \end{aligned} \)
\(\begin{aligned} \frac{I_A}{I_B} &= \frac{10^{4.5}}{10^{3.2}} \\[5pt] &= 10^{1.3} \\[5pt] &= 199.53 \ \text{times} \end{aligned} \)
Natural Logarithm
A natural logarithm is defined as a logarithm with base \(e\), yes, that wacky number from earlier. It is also noted with \(ln\):
\(\log_e{a} = \ln{a} \)
This logarithm can be utilised just as much as all other logarithms, and equally follows the log laws. It also has some interesting properties in calculus, which will be discussed later on.
7) Solve for \(x\):
a. \(e^{3x} \times e^{-1} = 6\)
\(\begin{aligned} \log_e{e^{3x-1}} &= \log_e{6} \\[5pt] \ln{e^{3x-1}} &=\ln{6} \\[5pt] 3x - 1 &= \ln{6} \\[5pt] 3x &= \ln{6} + 1 \\[5pt] x &= \frac{1}{3} \ln{6} \end{aligned} \)
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