Use the Product, Quotient, and Product Rules for Exponents to rewrite the following expression or solve the problem. Remember, the main goal is to exercise your brain, not to get the answers, so you must not use tools like Photomath or Mathway for anything but answer checking. Hint: Use one rule at a time. Don’t spend too much time worrying about which rule to use first! As soon as you see a rule that you could use, do that.
Use the integers 1 to 9 to fill in the boxes. There may be more than one correct answer! There is at least one solution that doesn’t use any digit more than once.
In Section 10.2, we learned how to write formulas for exponential models that increase by a constant percent change over awkward time periods. For example, in Activity 10.2.1, we discovered that if a bacteria colony starts initially weighs 3 g and doubles in size every 70 minutes, then a formula for its mass \(M\) after \(t\) minutes is
\begin{equation*}
M = 3 \times 2^{t/70}\text{.}
\end{equation*}
We can use exponent rules to rewrite this formula:
Use exponent rules to rewrite each of the formulas from Section 10.2, and use the rewritten formula to find the percent change per minute/hour/day/etc.
In Activity 10.2.2, we saw that if a new cat video is posted, 12 people view it the first day, every 4 days afterward the number of people who see it triples, and \(V\) is the number of views the video has after \(d\) days, then a formula for \(V\) is
\begin{equation*}
V = 12 \times 3^{t/4}\text{.}
\end{equation*}
In Activity 10.2.3, we learned that Barium-133 has a half-life of 10.551 years. If we start with 7.000 grams of Barium-133, then a formula for the amount of left, \(A\text{,}\) after \(t\) years is
\begin{equation*}
A = 7 \times \left(\frac{1}{2}\right)^{t/10.551}\text{.}
\end{equation*}
In Activity 10.2.4, we learned about the Chernobyl nuclear disaster. We were told that \(B = 14252(0.5)^{x/32}\) describes the amount, \(B\text{,}\) in becqurels, of a radioactive element remaining in the area \(x\) years after 1986.
In Activity 10.2.5, we learned that \(e \approx 2.71828\) is a special number used in science, finance, and other fields. Finance uses \(e\) to model continuously compounded interest with the formula
\begin{equation*}
A = Pe^{rt}
\end{equation*}
where \(A\) is the amount of money currently in an account, \(P\) is the amount of money initially deposited into the count, \(r\) is the interest rate as a decimal, and \(t\) is the number of years since the money was deposited. Suppose Parker deposits $1000 into an account that earns interest at a rate of 2% per year, compounded continuously. What is the equivalent interest rate if, instead of compounding continuously, we only compound once per year?
The relationship between logs and exponents mean that there are log properties that correspond to the Product, Quotient, and Power Rule of Exponents. They are:
