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Solving Systems of Equations
This unit introduces how to systematically solve a system of equations, namely linear equations. Examples of nonlinear systems, including systems of 3 unknowns will be of emphasis.

Graphs of Trigonometric Functions
The unit focuses primarily on how to graph periodic sinusoidal functions, and how to identify features of a waveform to produce an equation by inspection.

Polar Coordinate Functions
An introduction to the polar coordinate system.

Variation

Complex Numbers

Exponents and Radicals
This unit is an extension of what was introduced in Math 1131. To learn how to work with radicals, knowing your exponent laws in crucial. Hence, this unit begins with a thorough review.

Logarithmic Functions
This chapter introduces you to exponential functions, and how they can be solved using logarithms.

Trigonometric Identities and Equations

Analytic Geometry
Converting between exponential and logarithmic form
None of the math you’ve learned so far in this course and in the previous has gone so far to help us solve for ‘x’ in a situation depicted below. Notice that the variable is located in the exponent part of the equation:
${2}^{x}=8\phantom{\rule{0ex}{0ex}}$Through visual inspection, however, ‘x’ is equal to 3, but your math intuition will likely begin to fail once you start seeing decimals or larger numbers. Therefore, we need to rely on a method that more sustainable or something that will not fail. So for this and similar problems, we use logarithms. By making the base of the log 2, then finding the log of 8, we get x = 3:
${\mathrm{log}}_{2}\left(8\right)=x\phantom{\rule{0ex}{0ex}}x=3\phantom{\rule{0ex}{0ex}}$ This is read: “log base 2 of 8 equals x”. The main learning goal for this lesson is being able to convert from one form to another.
These two statements are completely equivalent, but what just happened? The base ‘2’ became the logarithm base, while the power ‘x’ became the answer. This relationship is summarized below, and you’ll need to know this for future reference.
⚠️ Before we continue, that as a reminder you were expected to have an Casio fx991ES Plus calculator for this course. This calculator, unlike most scientific calculators, has a feature that enables its users to customize the base of a logarithm. Rather, the general function found on most scientific calculators only allows you to find the answer when the base is 10. Therefore, for those that don’t have this calculator, these next few examples cannot be evaluated electronically, but you will be expected to know how to convert from exponential to logarithmic form. Instead, you’ll need to know the properties of logarithms that are being introduced next section to solve for the variable found in the exponent.
For the next two examples, remember the following:
${\mathrm{log}}_{{B}{a}{s}{e}}{A}{n}{s}{w}{e}{r}={E}{x}{p}{o}{n}{e}{n}{t}\iff {B}{a}{s}{{e}}^{{E}{x}{p}{o}{n}{e}{n}{t}}={A}{n}{s}{w}{e}{r}\phantom{\rule{0ex}{0ex}}$Convert the following equations to logarithmic form:
${5}^{3}=125$ $100={10}^{2}$ Answers:
${\mathrm{log}}_{5}\left(125\right)=3$ ${\mathrm{log}}_{10}\left(100\right)=2\phantom{\rule{0ex}{0ex}}$ By convention, when the base of the log is 10, we don’t write it in as it’s assumed:
$\therefore \mathrm{log}\left(100\right)=2$
Convert the following equations to exponential form:
${\mathrm{log}}_{8}2=x$ ${\mathrm{log}}_{5}\frac{1}{125}=x$ Solutions:
${8}^{x}=2\phantom{\rule{0ex}{0ex}}{\left({2}^{3}\right)}^{x}=2\phantom{\rule{0ex}{0ex}}{2}^{\overline{)3x}}={2}^{\overline{)1}}\phantom{\rule{0ex}{0ex}}3x=1\phantom{\rule{0ex}{0ex}}\overline{)x=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{1ex}{$3$}\right.}\phantom{\rule{0ex}{0ex}}$ ${5}^{x}=\frac{1}{125}\phantom{\rule{0ex}{0ex}}{5}^{x}=\frac{1}{{5}^{3}}\phantom{\rule{0ex}{0ex}}{5}^{\overline{)x}}={5}^{\overline{)\u20133}}\phantom{\rule{0ex}{0ex}}\overline{)x=\u20133}$
Always remember that the base of the exponent always becomes the base of the log, no matter what. Interesting though, there are two exceptions listed below:
 If the base is 10, you don’t write the base 10 as it’s assumed to be 10 when not written. These logs are referred to as common logs. A history note: common logarithms are also called Briggs’ logarithms, after Henry Briggs (1561–1630).
 If the base is e (euler’s number), rather than writing log_{e}, it gets its own symbol as ln (pronunciated like “lawn”).
Convert the exponential equations into logs:
$100={10}^{2}$ ${e}^{3}=x$ Answers:
${\mathrm{log}}_{10}100=2\phantom{\rule{0ex}{0ex}}\overline{)\mathrm{log}100=2}$ ${\mathrm{log}}_{e}x=3\phantom{\rule{0ex}{0ex}}\overline{)\mathrm{ln}x=3}$
Logarithms and Significant Figures
When a log is evaluated, let’s say:
$\mathrm{log}\left(17.56\right)=1.244524512\phantom{\rule{0ex}{0ex}}$The answer 1.244524512 consists of the whole number 1, called the characteristic, and the positive decimal number 0.2445…, called the mantissa. The mantissa is rounded based on the number of digits being logged; notice that 17.56 consists of 4 numbers, hence we round to four significant figures. In other words, the characteristic is never accounted for a significant figure.
$\therefore \mathrm{log}(17.56)=1.\overline{)2445}\phantom{\rule{0ex}{0ex}}$Logarithm Restrictions
Try calculating log(2) on your calculator. Doing so will result in a error. To explain why, set log(2) equal to x, and then convert it into an exponential equation:
$y=\mathrm{log}(\u20132)\phantom{\rule{0ex}{0ex}}{10}^{x}=\u20132\phantom{\rule{0ex}{0ex}}$The base is positive and is being raised to ‘x’. No matter what power you place into y, you will never get a negative output. This suggests that when you’re finding your restrictions for logarithmic functions, also remember that what you place into the log function must be greater than zero:
$y=\mathrm{log}\left(\underset{x>0}{\underset{\u23df}{x}}\right)\phantom{\rule{0ex}{0ex}}$Solving Basic Logarithmic Equations
Given what you’ve just learned about converting between logarithmic and exponential equations, let’s try solving the examples presented below:
Question 1
The temperature T of a bronze casting initially at 1250º F decreases exponentially with time. Its temperature after 5.00 s can be found from the equation:
$\mathrm{log}\left(\frac{T}{1250}\right)=\u20131.55\phantom{\rule{0ex}{0ex}}$Evaluate ‘T’.
SolutionStart by changing the equation to exponential form:
${10}^{\u20131.55}=\frac{y}{1250}\phantom{\rule{0ex}{0ex}}$Now solve for ‘y’:
$1250\xb7\left({10}^{\u20131.55}\right)=y\phantom{\rule{0ex}{0ex}}y=35.2\phantom{\rule{0ex}{0ex}}$The temperature after 5 seconds is 35.2 ºF.
[collapse]Question 2
A certain pendulum, after being released from a height of 4.00 cm, reaches a height y after 3.00 seconds, where y can be found from:
$\mathrm{ln}\left(\frac{y}{4.00}\right)=\u20130.50\phantom{\rule{0ex}{0ex}}$Evaluate ‘y’.
SolutionStart by changing the equation to exponential form:
${\mathrm{log}}_{e}\left(\frac{y}{4.00}\right)=\u20130.50\phantom{\rule{0ex}{0ex}}{e}^{\u20130.50}=\frac{y}{4.00}\phantom{\rule{0ex}{0ex}}4.00\xb7{e}^{\u20130.50}=y\phantom{\rule{0ex}{0ex}}y=2.43\phantom{\rule{0ex}{0ex}}$The height after 3 seconds is 2.43 cm.
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