Exponential Functions Practice Problems

Here you will find a concise set of exponential functions practice problems. Visit this page directly at hunkim.com/exponential

Which graph grows more quickly?
$y=x^2$ or $y=2^x$ or $y=x^{10/3}$
f(x)=2^x
1. Sketch
2. Domain?
3. Range?
4. Intercept(s)
5. $f(-3)?$
Sketch $y=\left(\frac{1}{3}\right)^x$
When does an exponential function experience growth vs decay?
Sketch $y=e^x$
Sketch $y=\pi^x$
Sketch $y=\frac{1}{2}4^x$
Sketch $y=-3(2)^{2x}$
Explain the variables in the exponential / geometric growth formula: $A=A_0 B^{t/p}$
T(x)=80(0.9)^x + 20
1. Sketch
2. Temperature is a function of time $x$ . Initial temperature?
3. How quickly does the temperature decrease?
4. Lowest possible temperature?
The population of a town changes by an exponential growth factor of $b$ every 5 years. If 2,000 people grows to be 10,000 in 15 years, find $b$ .
How many years does it take for your investment to triple in value if the annual interest rate is 10%?
$x^{2/3}=5$ . Solve $x$ .
Solve $5=3(2)^x$
Equation of the graph below?
A radioactive sample with an initial mass of 2mg has a half-life of 4 days. The equation that models the exponential decay, in 7-day intervals, is $A=2\left(\frac{1}{2}\right)^{kt}$ . Find $k$ .
A bacterial colony initially has 100 cells and triples each week. After $t$ days the population is modelled by the function $P(t)=100b^{t/k}$ . Find $b$ and $k$ .
Humor is contagious. On Day 1, you tell a joke to three friends. On Day 2, your friends tell three other friends each this joke. This pattern repeats. How many people will hear the joke on Day 4 only?
Explain the Compound Interest formula: $A=P\left(1+\frac{i}{n}\right)^{nt}$
You borrow $5000 on your credit card. The annual interest rate is 30%. If interest is compounded daily how much money do you owe if you do not make a payment for 24 months?
You buy a new car for $40,000. It depreciates by 20% each year. How much will your car be worth in 10 years?
Recall the compound interest formula: $A=P\left(1+\frac{r}{n}\right)^{nt}$ . Given the constant $e=\lim_{m\to\infty} \left(1+\frac{1}{m}\right)^m$ , why is the formula for continuously compounded interest $A=Pe^{rt}$ ?
Suppose you inherit and then invest $100,000. Interest is compounded continuously for 7 years at a rate of 10%. What is the new value of your inheritance?
$f(x)=10^x$ . Show that the inverse function is $y=\log_{10} x$

Answers

Growth: $x^2<x^{10/3}<2^x$
1. See $y=2^x$ below:
2. $x\in \reals$
3. $y>0$
4. $y=1$
5. $\frac{1}{8}=0.125$
See graph below:
Exponential growth when the base $b>1$ .
See $y=e^x$ below:
See $y=\pi^x$ below:
See $y=\frac{1}{2}4^x$ below:
See $y=-3(2)^{2x}$ below:
$A_0$ is the initial amount[/katex].
$B$ is the growth factor.
$t$ represents the time.
$p$ represents the period of time (think of the word “every” period of time)
a
1. See graph below:
2. 100 degrees
3. Loses 10% each increase in time $x$ .
4. 20 degrees
$b=\sqrt[3]5\approx1.71$
$\log_{1.1}3\approx11.5$ years
$5^{3/2}=5\sqrt{5}$
$x=\log_2\left(\frac{5}{3}\right)\approx0.737$
$y=4(3)^x+2$
$k=7/4$
$b=3$ and $k=3$
$3^4=81$
$P$ is the initial amount or Principal.
$i$ is the interest rate (write as a decimal number).
$n$ is the compounding period ex. compounded monthly means $n=12$ .
$t$ represents time (usually in years).
$5000\left(1+\frac{1.3}{365}\right)^{365\times2}$
$4,294.97
$A=P\left(1+\frac{r}{n}\right)^{nt}$ and $e=\lim_{m\to\infty} \left(1+\frac{1}{m}\right)^m$
Let $n=mr$
$\lim_{n\to\infty}P\left(1+\frac{r}{n}\right)^{nt}$
$=\lim_{m\to\infty}P\left(1+\frac{1}{m}\right)^{mrt}$
$P{\left( {\mathop {\lim }\limits_{x \to \infty } {{\left( {1 + \frac{1}{m}} \right)}^m}} \right)^{rt}} = P{e^{rt}}$
$174,871.71
To find the inverse function begin by replacing x with y and y with x.

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