## Properties of the Exponential functions

For x and y real numbers:

- a
^{x} a ^{y} = a ^{x + y}

example: 2 ^{3} 2^{5} = 2 ^{8}

- (a
^{x}) ^{y} = a ^{x y}

example: (4 ^{2}) ^{5} = 4 ^{10}

- (a b)
^{x} = a ^{x} b ^{x}

example: (3 × 7)^{3} = 3^{3} 7^{3}

- (a / b)
^{x} = a ^{x} / b ^{x}

example: (3 / 5)^{3} = 3 ^{3} / 5 ^{3}

- a
^{x} / a ^{y} = a ^{x - y}

example: 5 ^{7} / 5 ^{4} = 5 ^{3}

## Questions with Detailed Solutions and Explanations

__Question 1__

Simplify the following expression
2 ^{x} - 2 ^{x + 1}
Solution to Question 1
- Use property (1) above to write the term 2
^{x + 1} as 2 ^{x} × 2 in the given expression

2 ^{x} - 2 ^{x + 1} = 2 ^{x} - 2 ^{x} × 2

- Factor 2
^{x} out

2 ^{x} - 2 ^{x + 1} = 2 ^{x}(1 - 2)

- Simplify to obtain

2 ^{x} - 2 ^{x + 1} = - 2 ^{x}

__Question 2__

Find parameters A and k so that f(1) = 1 and f(2) = 2, where f is an exponential function given by

f(x) = A e ^{k x}

Solution to Question 2
- Use the fact that f(1) = 1 to obtain

1 = A e ^{k}

- Now use f(2) = 2 to obtain

2 = A e ^{2 k}

- Rewrite the above equation as

2 = A e^{k} e^{k}

- Use the first equation 1 = A e
^{k} obtained in the first step to rewrite 2 = A e^{k} e^{k} as

2 = e ^{k}

- Take the ln of both sides and simplify to obtain

k = ln (2)

- To obtain parameter A, substitute the value of k obtained in the equation 1 = A e
^{k}.

1 = A e ^{ln(2)}

- Simplify and solve for A.

A = 1/2

- Function f is given by

f(x) = (1/2) e ^{x ln(2)}

- Which can be written as

f(x) = (1/2) (e ^{ln(2)})^{ x}

- and simplified to

f(x) = 2 ^{x - 1}

__Check answer against given information__

f(1) = 2 ^{1 - 1}

= 1

f(2) = 2 ^{2 - 1}

= 2

__Question 3__

The populations of 2 cities grow according to the exponential functions

P1(t) = 100 e ^{0.013 t}
P2(t) = 110 e ^{0.008 t}

where P1 and P2 are the populations (in thousands) of cities A and B respectively; t is the time in years such that t is positive and t = 0 corresponds to the year 2004.

When will the populations of the two cities be equal and what will be their populations?
Solution to Question 3

- Let t = t' be the time when P1 and P2 are equal, this leads to the following equation in t'

100 e ^{0.013 t'} = 110 e ^{0.008 t'}

- Divide both side of the above equation by 100 e
^{0.008 t'} and simplify to obtain

e ^{0.013 t'} / e ^{0.008 t'} = 110/100

- Use property of exponential functions a
^{x} / a^{y} = a ^{x - y} and simplify 110/100 to rewrite the above equation as follows

e ^{0.013 t'- 0.008 t'} = 1.1

- Simplify the exponent in the left side

e ^{0.005 t'} = 1.1

- Rewrite the above in logarithmic form (or take the ln of both sides) to obtain

0.005 t' = ln 1.1

- Solve for t' and round the answer to the nearest unit.

t' = (ln 1.1) / 0.005.

t' is approximately equal to 19 years.

the year will be: 2004 + 19 = 2023.

- Find the populations when t = t' = 19 years. Use any of the function P1 or P2 since they are equal at t = t'

P1(t') = 100 e ^{0.013*19}

P1(t') is approximately equal to 128 thousands.

For checking, the graphical solution to the above problem is shown below.

__Question 4__

The amount A of a radioactive substance decays according to the exponential function

A(t) = A _{0} e ^{r t}

where A_{0} is the initial amount (at t = 0) and t is the time in days (t ? 0). Find r, to three decimal places, if the half life of this radioactive substance is 10 days.
Solution to Question 4

- At t = 10 days, the amount A of the substance would be equal to half
the initial amount A
_{0} (definition of half life)

A _{0} e ^{r×10} = A _{0} / 2

- Divide both side of the above equation by A
_{0}

e ^{10 r} = 1 / 2

- Rewrite the above equation in logarithmic form (or take ln of both sides) to obtain

10 r = ln (1/2)

- Solve for r

r = 0.1 ln(1/2)

- Approximate r to 3 decimal places.

r is approximately equal to -0.069.

For checking, the graph of A(t) = 100 e^{-0.069t} is shown below.

Note at t = 0 A = 100 (initial amount) and at t = 10 (half life), A is approximately equal to 50 which half the initial amount 100.

## More Questions with Answers