Exponential Functions Questions with Solutions

Properties of the Exponential functions

For x and y real numbers:

1. a ^x a ^y = a ^{x + y}
   example: 2 ³ 2⁵ = 2 ⁸
   
2. (a ^x) ^y = a ^{x y}
   example: (4 ²) ⁵ = 4 ¹⁰
   
3. (a b) ^x = a ^x b ^x
   example: (3 × 7)³ = 3³ 7³
   
4. (a / b)^x = a ^x / b ^x
   example: (3 / 5)³ = 3 ³ / 5 ³
   
5. a ^x / a ^y = a ^{x - y}
   example: 5 ⁷ / 5 ⁴ = 5 ³

Questions with Detailed Solutions and Explanations

• 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

• 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

• 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

• At t = 10 days, the amount A of the substance would be equal to half the initial amount A₀ (definition of half life)
  A ₀ e ^r×10 = A ₀ / 2
  
• Divide both side of the above equation by A₀
  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