The Log Operation Prep Activities

Section 8.1 The Log Operation Prep Activities

Prep Activity 8.1.1.

Let’s consider the steps we’d take to solve several equations!

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(a)

Fill in the blanks:

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In order to solve the equation \(x+7= -2\) you could from both sides of the equation.

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This works because the operation of "undoes" the operation of .

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After performing this step, you arrive at the solution, \(x = \fillinmath{XXXXXXXXXX}\text{.}\)

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(b)

Fill in the blanks:

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In order to solve the equation \(3x = 15\) you could both sides of the equation by .

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This works because the operation of “undoes” the operation of .

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After performing this step, you arrive at the solution, \(x = \fillinmath{XXXXXXXXXX}\text{.}\)

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(c)

Fill in the blanks:

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In order to solve the equation \(x^2 = 15\) you could both sides of the equation.

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This works because the operation of “undoes” the operation of .

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After performing this step, you arrive at the solution, \(x = \fillinmath{XXXXX} \text{ or } \fillinmath{XXXXX}\text{.}\)

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Now we would like to introduce a new operation which “undoes” exponentiation. That operation is written as \(\log(x)\text{,}\) and we say this as “the log of x”.

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The term log is an abbreviation for the mathematical term logarithm. A logarithm is a quantity representing the power to which a fixed number (the base) must be raised to produce a given number.

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For example, a logarithm, or log operation, can be used to find the exponent to which one must raise the number 10 to get the specific number as the output.

The expression “\(\log(1000)\)” is then equal to \(3\text{,}\) since \(10^3=1000\text{.}\)

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The expression “\(\log(10)\)” is then equal to \(1\text{,}\) since \(10^1=10\text{.}\)

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The expression “\(\log\left(10^x\right)\)” is then equal to \(x\text{,}\) since \(10^x=10^x\text{.}\)

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Most scientific and graphing calculators have a button or function labeled “log” which performs this calculation. Calculators will assume that the base is 10 unless you specifically ask them to use a different base. We’ll stick with base 10 for now.

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Prep Activity 8.1.2.

What are the following logs? Try to answer without using a calculator!

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(a)

\(\log(100) = \fillinmath{XXXXX}\)

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(b)

\(\log(100,\!000) = \fillinmath{XXXXX}\)

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Now we would like to use the log operation to solve equations. Consider the equation \(10^x=16\text{.}\) Notice that the exponent is not a whole number in this case. We can still solve this equation by doing something to both sides of the equation, as we did in Prep Activity 8.1.1. In this case, we will “take the log” of both sides of the equation:

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Table 8.1.1. Solving an Equation with a Log

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Initial equation	\(10^x = 16\)
Take the log of both sides	\(\log\left(10^x\right) = \log(16)\) \(x = \log(16)\)
Use a calculator’s “log” function to find \(\log(16)\) and get the answer	\(x \approx 1.204\)

We can can check this result using a calculator. If \(x = 1.204\text{,}\) we get

\begin{equation*} 10^x= 10^{1.204} = 15.996\text{,} \end{equation*}

which is about \(16\text{.}\) It is not exactly \(16\) because we rounded \(\log 16\) to three decimal places. Logs are often terrible decimals.

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Prep Activity 8.1.3.

(a)

Using your calculator, follow the steps above to solve the equation \(10^x = 25\text{.}\) Round to the nearest thousandth.

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(b)

Check your answer by finding \(10^x\text{,}\) where \(x\) is your answer. You should get \(25\text{,}\) or something very close. Did you?

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(c)

What happens if you try to follow the steps to solve the equation \(10^x = -25\text{?}\) Why do you think that is?

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Prep Activity 8.1.4.

You’ll need to be able to do the following things for this lesson. Rate how confident you are on a scale of 1 - 5 (1 = not confident and 5 = very confident).

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Table 8.1.2.

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Skill or Concept: I can …	Rating from 1 to 5
Solve equations by undoing operations.
Understand that the log operation undoes the “raise 10 to the power of...” operation.

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