Getting started
Master the fundamentals of exponential, logarithmic, hyperbolic, and parametric equations.
Logarithms
The logarithm is the inverse of the exponential.
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You can read \(\log_b(x)\) as “\(b\) raised to what power equals \(x\)?”
{% include info.html text=“The log base 10 of a number (rounded down) is 1 less than the number of digits of the number” %}
For example;
\(\log_{10}(1000) = 3\)
\(\log_{10}(49323) \approx 4.69 \rightarrow\left \lfloor{\log_{10}(49323)}\right \rfloor = 4\)
\(\log_{10}(333333) \approx 5.52 \rightarrow\left \lfloor{\log_{10}(333333)}\right \rfloor = 5\)
{% include info.html text=“In general, every positive number with 1 followed by only 00s will have an integer answer when taking the logarithm. Adding a 0 will ‘step’ to the next integer.” %}
\(\log_{10}100=2 \;\; \log_{10}1000=3 \;\; \log_{10}10000=4\)
With a logarithmic scale, moving forward a step multiplies the distance by a set amount. So For base 10; the scale, started a 0, i.e. $10^0 = 1. Each step is akin to multiplying the distance by 10. Moving right a linear amount \(x\) causes us to multiply the distance by \(10^x\).
\(10^2 * 10^3 = 10 ^ {(log_{10}(10^2) + log_{10}(10^3))}\)
\(\log_{10}(ab) = log_{10}(a) + log_{10}(b)\)
Exponential equations
The rate of growth is proportional to the current amount at any given time.
\[ y = a + b^x \]
It is impossible to undo the effect of doubling \(b\) with a change in \(a\)
Algebraically, if we start off with \(y = a \times b^xy=a×b^x\) then doubling \(b\) gives \(y = a \times (2b)^x = a \times 2^x \times b^x\).
So each point is being multiplied by an additional \(2^x\) term.
This can’t be undone by changing \(a\) because the value of \(2^x\) changes as \(x\) changes while the value of \(a\) will remain constant.
So while it is possible to change \(a\) so that the effect of doubling \(b\) is undone for a single point it will not work for all other points.
Changing the base
\[ y = b^x \]
For all bases where \(b > 0\), all exponential functions pass through the point [0, 1]
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As \(b\) grows, the proportion of the graph for \(x > 0\) becomes steeper, leading to a right angle shape.
The graph can be reflected over the y axis when \(b\) is below 1. \(y = 0.5^x\) reflects \(y = 2^x\)
{% include info.html text=“An exponential function can’t be defined with a negative base” %}
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Changing \(b\) is the only thing that will change the shape of the curve. The others shift or scale the graph.
Exponential arithmetic
\[ x^2 x^3 x^4 = x^9 \]
This is an example of the product rule.
{% include info.html text=“these rules are true if \(a\) is positive, and \(m\) and \(n\) are real numbers.” %}
If we have an expontential equation, \(y = ab^{x +c}\), only changing \(b\) will change the shape of the curve. Changing \(x\) or \(c\) changes where the curve falls/steepens.
An implication of the product rule for exponential expressions;
muliplication or division in one part of the expression can have the same effect as addition or subtraction in another part of the expression
compressing or expanding the graph vertically (multiplying or dividing the entire expression) has the same effect as shifting the graph horizontally (adding or subtracting from the exponent).
Increasing \(c\) stretches the graph vertically, increasing \(a\), by a larger factor, will have the same effect.
\[ y = 9 \times 3^x \]
Can be rewritten using the product rule
\[ y = 3^2 \times 3^2 = 3^{x +2} \]
Exponential functions can be rewritten as logarithmic functions
\[ b^a = c \equiv log_b(c) = a \]
Graphing Logarithms
All basic logarithmic function have the same general shape.
Asking the value of \(log_{10}(x)\) is equivalent to asking “what value exponeont does 10 need to be raised to in order to get \(x\)?
Think of the inverse function, \(log_{10}(x) \equiv y = 10^x\)
The domain of \(log_{10}(x)\) will be equal to the range of \(y = 10^x\)
Domain being the complete set of possible values of x.
All basic loagrithmic functions of the form \(f(x) = log_b(x)\) have an “anchor point” - a point that they all pass through, regardless of their base.
Regardless of the base, \(b\), we always have \(log_b(1) = 0\) because \(b^0 = 1\) for any \(b\). That means all logarithmic functions will pass through the point where x = 1 and y = 0, or (1,0)
{% include info.html text=“If \(x\) is less than 1, \(x\) we need to be raised to a negative exponent.
For example, with base 10, \(10^{-3} = \frac{1}{10^3}\)
As \(x\) gets larger, the rate at which \(\log_{10}(x)\) increases will become slower and slower” %}
The graph of \(y = \log_{10}(x)\) is a reflection of the graph \(y = 1-^x\) over the line \(y=x\). This is a result of the inverse relationship between the two functions.
It is not possible to define \(\log_1(x)\) because \(y = 1^x\) has the same output for any value of \(x\), \(1^{60}\), is the same as \(1^{900}\).
Understanding logarithmic arithmetic
\[ 10^5 - 10^2 \approxeq 10^5\]
\[ 10^5 - 10^2 = 99900 = 10^{\log_{10}(99900)} \approxeq 10^{4.999} \]
\[ \log_{10}10^Q = Q \]
Say you have a logarthimic scale.
Adding logarithms is equivalent to adding linear distances on the chart, and the values inside the logarithms (the distances in space) will get multiplied.
If we a distance of $(s) $ and then move an additional ${10}(t) $ we have moved ${10}st $
\[ \log_{10}s + log_{10}t = \log_{10}st \]
This is how much we have moved on the chart, not through space, so you’re getting the point on the axis.
\[ \log_{10}10^2 + \log_{10}10^2 + \log_{10}10^2 = \log_{10}((10^2)(10^2)(10^2)) = \log_{10}(10^6) \]
If we move a distance of $_{10}(m) $ and repeat is \(p\) times, so \(p \times \log_{10}(m)\) we get \(log_{10}(m^p)\)
In the example, \(p = 3\), so we have \((10^2)^3 = (10^6)\), not \((10^2 \times 3)\) which is \((m \times p)\)
Take the inverse
\[ \log_{10}n - \log_{10}d = \log_{10}\frac{n}{d}\]
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\[ \log_b(s) + \log_b(t) = \log_b(st) \] \[ \log_b(n) - \log_b(d) = \log_b(\frac{n}{d}) \] \[ (p)\log_b(m) = \log_b(m^p) \]
\[ \log _{3}(x)+\log _{3}(y)-\log _{3}(z) = \log _{3}\left(\frac{x y}{z}\right)\]
Say we have;
\[y = \log{b}(ax) + c\]
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Practice
- Express the following in a single log
\[ \log _{2}(x)+\log _{2}(x)+\log _{2}(x)+\log _{2}(x) \]
\[\log _{2}(x \cdot x \cdot x \cdot x)=\log _{2}\left(x^{4}\right) \]
- Solve for \(x\) : \[ \log _{8}(x)-\log _{8}(4)=\log _{8}(36)-\log _{8}(x) \]
\[ \begin{aligned} \log _{8}(x)-\log _{8}(4) &=\log _{8}(36)-\log _{8}(x) \\ \log _{8}\left(\frac{x}{4}\right) &=\log _{8}\left(\frac{36}{x}\right) \\ \left(\frac{x}{4}\right) &=\left(\frac{36}{x}\right) \\ x^{2} &=144 \\ x &=12 \end{aligned} \]
- Note
$$_{b}(m) = y b^y = m $£
We also now that
\[\log _{z}\left(b^{y}\right)=\log _{z}(m) \]
- If you now isolate \(y\), what does it equal?
Applying the rule \((p) \log _{b}(m)=\log _{b}\left(m^{p}\right)\) \[ \begin{aligned} \log _{z}\left(b^{y}\right) &=\log _{z}(m) \\ (y) \log _{z}(b) &=\log _{z}(m) \\ y &=\frac{\log _{z}(m)}{\log _{z}(b)} \end{aligned} \]
Change of base
Change of base is a very useful procedure to manipulate log functions into something more useful.
\[ \log _{a}(b)=\frac{\log _{c}(b)}{\log _{c}(a)} \]
\(\text { If } \log _{9}(243)=2.5, \text { what is the value of } \log _{3}(243) ?\)
\[ \log _{9}(243)=\frac{\log _{3}(243)}{\log _{3}(9)} = 2.5 \\ \frac{\log _{3}(243)}{2} = 2.5 \\ \log _{3}(243)= 5 \\ \]
Simplify;
\[ \frac{\log _{2}(a)}{\log _{4}(a)} \]
Instincitvely you know to are going to have to raise 4 by a power \(x\) fewer times to reach \(a\) than 2. That means \(\log _{2}(a) > \log _{4}(a)\)
We can apply the change of base formula to both the numerator and denominator. The new base we choose doesn’t matter, so we will just use 10: \[ \log _{2}(a)=\frac{\log _{10}(a)}{\log _{10}(2)}, \quad \log _{4}(a)=\frac{\log _{10}(a)}{\log _{10}(4)} \] Substituting back into our starting expression, we now have \[ \frac{\frac{\log _{10}(a)}{\log _{10}(2)}}{\frac{\log _{10}(a)}{\log _{10}(4)}}=\frac{\log _{10}(4)}{\log _{10}(2)} \] Now \(4=2^{2}\), so \(\log _{10}(4)=\log _{10}\left(2^{2}\right)=2 \log _{10}(2)\), so we can rewrite the expression above as \[ \frac{2 \log _{10}(2)}{\log _{10}(2)}=2 \]
Logarithmic equations
{% include info.html text=“When solving equations with logarithms, a general strategy is to rewrite the equation in exponential form” %}
\[\log _{a}(x)=y \text { is equivalent to } a^{y}=x \]
Solve for \(x\):
- \[\log _{5}(x+1)=2 \]
\[ 5^2 = x+ 1 = 25\]
- \[\log _{10}(3x+1)=2 \]
\[ 10^2 = 3x+ 1 = 100\]
- \[\log _{4}(x) +log_{4}(x+6)=2 \]
\[ \log_{4}(x^2+6x)=2 \]
\[ 4^2=x^2 + 6x\]
\[ x^2 + 6x - 16 = 0\]
Using this along, we could consider \(x\) to equal 2 or -8
\[ (x+8)(x-2)=0\]
However \(\log _{4}(x) \text{ can't be defined for } x=-8\)
\[ \left(\log _{4}(x)\right)^{2}+\log _{4}\left(x^{3}\right)-4=0 \]
Retwrite \(log _{4}(x^{3})\) as \(3\log_{4}(x)\)
Substitute \(\log_{4}(x)\) for \(y\)
\[ y^2 + 3y - 4 = 0 \]
\[ (y + 4 )(y - 1) = 0\]
\[ log_{4}(x) = -4 \text{ or } 1 \]
\[x=4^{-4}=\frac{1}{256} \quad \text { or } \quad x=4^{1}=4\]
\[\log _{2}(x)+\log _{4}(9)=\log _{2}(12)\]
Change base \(\log _{4}(9) = \frac{log_2(9)} {log_2(4)}\)
\[ log_2(x) + 1.5849625007211563 = 3.5849625007211565 \]
\[ log_2(x) = 2 \]
Note, could have rewritten the rebasing as:
\[ \frac{log_2(9)} {log_2(4)} = \frac{\log _{2}(9)}{2}=\left(\frac{1}{2}\right)\left(\log _{2}(9)\right)=\log _{2}\left(9^{\frac{1}{2}}\right)=\log _{2}(3) \]