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.
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