Exponents & Logarithms

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Exp & Log functions

Discussion

Stewart EJ, Madden R, Paul G, Taddei F (2005), CC BY-SA 4.0

The video above shows bacteria splitting. Starting with one bacterium, the population doubles every 36 minutes.

Find f(t), the number of bacteria as a function of time, t, in minutes.

We want a function f(t) that gives the number of bacteria after t minutes, knowing the population “doubles” every 36 min. The population will be:

1 after 0 min

2 after 36 min

4 after 72 min

8 after 108 min

and so on.

An exponential with base 2 does exactly that: after n doublings you have 2n times the start.

Since we start with one bacterium, after n doublings there are

f=1×2n=2n

Next, we must express n, the number of doublings, in terms of time t. Every doubling takes 36 minutes, so in t minutes the number of doublings is

n=36t

Hence

f(t)=2t/36

Check:

f(0)=20=1,f(36)=236/36=2

so at t=36 min the population has indeed doubled.

Discussion

Consider the function g(t)=2t.

Fill in the table below.
t
−5
−1
0
1
2
3
g(t)
Then, plot those points on the grid below.
What are some key features of the graph?

Part (a)

We have g(t)=2t. For each given t:

g(−5)=2−5=251=321

g(−1)=2−1=21

g(0)=20=1

g(1)=21=2

g(2)=22=4

g(3)=23=8

Filling these into the table:

t	−5	−1	0	1	2	3
g(t)	321	21	1	2	4	8

Part (b)

From part (a) we have the points

(−5,321),(−1,21),(0,1),(1,2),(2,4),(3,8),

wich are plotted below.

Now, when we draw a smooth curve through them, we see the curve approaches the horizontal line y=0 as t→−∞ and rises ever more steeply for large positive t.

Part (c)

The graph of g(t)=2t has three key features:

y-intercept: Since g(0)=20=1, the curve crosses the y-axis at (0,1). Visually you’ll see a point on the y-axis one unit above the origin.
Horizontal asymptote: As t→−∞, 2t becomes very small but remains positive, approaching 0 without ever reaching or going below it. Thus the line
y=0
is a horizontal asymptote: to the left, the line gets increasingly close to this line but never touches it.
Exponential growth: Each time t increases by 1, g(t) doubles. For larger t the curve rises more and more steeply, so the graph turns sharply upward on the right, giving the characteristic “J” shape.
Domain and Range: the domain of the function is (−∞,∞) since 2t is defined for all t. The range of the function is (0,∞) since 2t is always positive even though it gets extremely close to 0 for large negative t.

Generalizing from our example of g(t)=2t, we see that an exponential function's graph is largely defined by:

Its y-intercept.
Falling toward - but never passing - its horizontal asymptote, which restricts range.
Increasing faster and faster toward infinity.

Exponential functions

An exponential function has the form f(x)=ax for some base a>0. The domain of f is R, and the range is f(x)>0:

However, exponential curves can have several transformations applied to them.

Consider the function f(x)=2⋅3x−1. What will its graph look like?

As always for a function, the y-intercept is at f(0)=2⋅30−1=2⋅1−1=1.
We also know that 3x is always positive, but for very large negative x 3x→0. So f(x) will have a horizontal asymptote at y=−1.
And since 3x is exponential, we will see the function increasing faster and faster to the right.

Graphing Exponential Functions

In general, to graph an exponential function of the form f(x)=cax+k, find the y-intercept of the curve, then analyze the behavior of the function on both ends (as x→∞ and as x→−∞). If possible, plotting other easily calculated points - often f(1) or f(−1).

The y-intercept is at (0,c+k) because f(0)=ca0+k=c(1)+k.
On one end, the curve will approach y=k.
- For a<1, as x→∞, f(x)→c(0)+k.
- For a>1, as x→−∞, f(x)→c(0)+k.
On the other end, the curve will rise with increasing steepness.

Exercise

The following graph shows the curve y=f(x).

Find the equation of f(x).

Select the correct option

Exponential growth

Exponential growth describes quantities that increase by the same factor over a certain amount of time. Algebraically, exponential growth are functions of the form

f(t)=Abt,

where b>1. b is called the growth factor.

Note: Aekt is another model for exponential growth if the instantaneous growth rate, k, is positive.

Stewart EJ, Madden R, Paul G, Taddei F (2005), CC BY-SA 4.0

Checkpoint

f(x)=10⋅3x. Find f(9)f(10).

Select the correct option

Discussion

Explain why 3⋅2x+3=24⋅2x.

Discussion

Some quantities get smaller over time rather than larger. If g(t)=xt is a function of exponential decay, what two numbers do you think x should fall between?

We want g(t)=xt to decrease as t increases, so for each t

g(t+1)<g(t).

Substitute g:

xt+1<xt.

Since for decay we also want g(t) to stay positive (we want to have less of something, not negative of it), xt>0. Dividing both sides by xt gives

x<1.

To avoid alternating signs (as happens if x is negative) we require

x>0.

Hence the only way to have a positive quantity xt that gets smaller with t is

0<x<1.

We can check with examples:

If x=2, then 2t grows as t increases.
If x=0.5, then 0.5t decreases toward 0.
If x=−0.5, the sign alternates, so it isn’t a simple decay.

Therefore x must lie between 0 and 1.

Exponential decay

Exponential decay describes quantities that decrease by the same factor over a certain amount of time. Exponential decay are functions of the form

f(t)=Abt,

where 0<b<1. b is called the decay factor.

Note: Aekt is another model for exponential growth if the instantaneous growth rate, k, is negative.

A common subject for IB questions about exponential decay is radioactive half-lives. A half-life is the amount of time a radioactive material takes to halve in quantity. These are a popular topic for exponential decay questions because every radioactive half-life can be modeled by a base 21 exponential!

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Checkpoint

The graph shows the equation y=3(43)t. Find t when y=1627.

Select the correct option

Discussion

Stewart EJ, Madden R, Paul G, Taddei F (2005), CC BY-SA 4.0

Previously, we found that f(t)=236t models the number of bacteria as a function of time. Can you think of a function f−1(x) that gives the time as a function of the number of bacteria?

We want the time t (in minutes) needed to reach x bacteria. Since the population doubles every 36 minutes, write:

Let the number of doublings be N. Then after N doublings starting from 1 bacterium we have

x=2N

and by definition N is the number for which 2N=x. We call this number

N=log2(x),

since log2(x) is “the exponent to which 2 must be raised to give x.” Finally, each doubling takes 36 minutes, so

t=36N=36log2(x)

is the time in minutes for the population to reach x.

Hence, f−1(x)=36log2(x).

Logarithmic functions

A logarithmic function has the form f(x)=logax, for a>0. The domain of f is x>0, and the range is R:

Exercise

Consider the function f(x)=4+log5(25−x2).

Find

the domain of f,
the range of f.

Select the correct option

Discussion

Ally argues that the inverse of the function f(x)=3x is a cube root. Is she right? Provide algebraic reasoning to support your answer.

To test Ally’s claim that f−1(x)=3√x, check the composition

f−1(f(x))=(3x)1/3=3x/3

which is not equal to x in general.

For example, if x=6:

f(6)=36=729,3√729=9=6

so cube root fails to “undo” 3x.

Instead, the inverse must satisfy f−1(f(x))=x. To reverse “raise 3 to the power …,” we ask “which exponent of 3 gives this number?” That operation is log3. Indeed, if we define

f−1(x)=log3(x)

then

f−1(f(x))=log3(3x)=x

and likewise

f(f−1(x))=3log3(x)=x

Answer: Ally is not right. The inverse of f(x)=3x is

f−1(x)=log3(x),

not a cube root.

Log and exponent functions are inverses

The functions logax and ax are inverses:

loga(ax)=x,alogax=x

This can be seen by the symmetry of their graphs in the line y=x:

Exercise

Consider the function f(x)=7⋅2x−9.

Find f−1(x).

Select the correct option

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