College Algebra Problems With Answers
sample 3 : Exponential and Logarithmic Functions

College algebra problems on logarithmic and exponential function with answers, are presented along with solutions are at the bottom of the page.

Problems

  1. Let the logarithmic function \( f \) be defined by \( f(x) = 2\ln(2x - 1) \).
    a) Find the domain of \( f \).
    b) Find vertical asymptote of the graph of \( f \).

  2. Let the exponential function \( h \) be defined by \( h(x) = 2 + e^x \).
    a) Find the range of \( h \).
    b) Find the horizontal asymptote of the graph of \( h \).

  3. The population of city A changes according to the exponential function
    \( A(t) = 2.9 \times 2^{0.11t} \) (millions)

    and the population of city B changes according to the exponential function
    \( B(t) = 1.7 \times 2^{0.17t} \) (millions)

    where \( t = 0 \) correspond to 2009.
    a) Which city had larger population in 2009?
    b) When will the sizes of the populations of the two cities be equal?

  4. Find the inverse of the logarithmic function \( f \) defined by \( f(x) = 2 \log_5(2x - 8) + 3 \).

  5. Find the inverse of the exponential function \( h \) defined by \( h(x) = -2 \cdot 3^{-3x + 9} - 4 \).

  6. Solve the logarithmic equation defined by
    \( \ln(2x - 2) + \ln(4x - 3) = 2 \ln(2x) \)


  7. \( A \), \( B \) and \( k \) in the exponential function \( f \) given by
    \( f(x) = Ae^{kx} + B \)

    are constants. Find \( A \), \( B \) and \( k \) if \( f(0) = 1 \) and \( f(1) = 2 \) and the graph of \( f \) has a horizontal asymptote \( y = -4 \).

Solutions to the Above Problems
    1. solve \( 2x - 1 > 0 \) to find domain: \( x > \dfrac{1}{2} \)
    2. solve \( 2x - 1 = 0 \) to find vertical asymptote: \( x = \dfrac{1}{2} \)

    1. range of \( h \): \( (2, +\infty) \)
    2. horizontal asymptote: \( y = 2 \)

    1. \( A(0) = 2.9 \) millions , \( B(0) = 1.7 \) millions, city A had larger population.
    2. solve \( 2.9 \times 2^{0.11t} = 1.7 \times 2^{0.17t} \), to find \( t \).
      take \( \ln \) of both sides of the equation
      \( \ln[ 2.9 \times 2^{0.11t} ] = \ln[ 1.7 \times 2^{0.17t} ] \)
      \( \ln(2.9) + 0.11t \ln(2) = \ln(1.7) + 0.17t \ln(2) \)
      solve for \( t \):
      \( t = \dfrac{\ln1.7 - \ln2.9}{0.11\ln2 - 0.17\ln2} = 13 \) (approximated to the nearest unit)
      The size of the two populations will be the same in 2009 + 13 = 2022.

    4. solve the equation: \( x = 2 \log_5(2y - 8) + 3 \) for \( y \) to obtain the inverse of the function.
      \( f^{-1}(x) = \dfrac{1}{2} 5^{ \dfrac{x-3}{2} } + 4 \)

    5. solve the equation: \( x = -2 \cdot 3^{-3y + 9} - 4 \) for \( y \) to obtain the inverse of the function.
      \( h^{-1}(x) = \dfrac {-1}{3}\log_3 \left( \dfrac{x+4}{-2} \right) + 3 \)

    6. Rewrite the given equation as follows
      \( \ln(2x - 2)(4x - 3) = \ln(2x)^2 \)

      The above gives the algebraic equation
      \( (2x - 2)(4x - 3) = (2x)^2 \)

      Solve the above quadratic equation for \( x \)
      \( x = 3 \) and \( x = \dfrac{1}{2} \)

      Check the two values of \( x \) and only \( x = 3 \) is a solution to the given equation.

    7. Horizontal asymptote \( y = -4 \) gives \( B = -4 \).
      \( f(0) = A + B = 1 \)

      which gives \( A = 5 \) since \( B = -4 \)
      \( f(1) = 5e^k - 4 = 2 \)

      solve for \( k \) to obtain: \( k = \ln\left(\dfrac{6}{5}\right) \)

    More References and links

    logarithmic functions
    exponential functions
    Algebra Questions and problems
    More ACT, SAT and Compass practice

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