Honors Algebra 2: Unit Review Problems

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  1. Solve: \( \displaystyle x - xe^{5x + 2} = 0 \)
    2. Solution

    \( \displaystyle x \in \left\{ - \frac{2}{5}, 0 \right\} \)

  2. Solve: \( \displaystyle \log_2{\left( x^2 - x - 6 \right)} + \log_{0.5} {(x - 3)} < 2\log_2{3}\)
    3. Solution

    \( \displaystyle 3 \lt x \lt 7 \)

  3. Solve: \( \displaystyle \log_{\frac{1}{3}}{(3x - 1)} - \log_{\frac{1}{3}}{(x + 1)} \gt 1 \)
    4. Solution

    \( \displaystyle \frac{1}{3} \lt x \lt \frac{1}{2} \)

  4. Solve: \( \displaystyle \log_{0.3}{(10x + 3)} < \log_{0.3}{(7x-4)} \)
    5. Solution

    \( \displaystyle x \gt \frac{4}{7} \)

  5. Write as a single, simplified logarithm:: \( \displaystyle \ln { \left( \frac{x}{x^2 - 1} \right)} + \ln { \left( \frac{x^2 + 2x + 1}{3x} \right)} \)
    6. Solution

    \( \displaystyle \ln{\frac{x+1}{3x-3}} \)

  6. Given the graphs of the functions \( \displaystyle f(x) = e^x \) and \( \displaystyle g(x) = e^{-x}\).
    1. What are the coordinates of the point(s) of intersection of \( y = f(x) \) and \( y = g(x) \)?
      Solution

      \( \displaystyle (0, 1) \)

    2. Suppose \( \displaystyle f(x) \) was shifted 3 units to the left, and \( g(x) \) was shifted three units to the right. What would be the coordinates of the new point of intersection?
      3. Solution

      \( \displaystyle (0, e^3) \)

    3. Suppose \( \displaystyle f(x) \) was stretched horizontally by a factor of 2 and translated 3 units to the left, while \( g(x) \) was compressed horizontally by a factor of 2 and shifted 2 units to the left. What would be the coordinates of the new point of intersection?
      4. Solution

      \( \displaystyle \left( - \frac{11}{5}, e^{\frac{2}{5}} \right) \)

  7. Show: \( \displaystyle \log_5{x} + \log_8{x} = \frac{\ln{40}}{\ln{5} \cdot \ln{8}} \cdot \ln{x} \)
    8. Solution

    \( \displaystyle \log_5{x} + \log_8{x} = \) \( \displaystyle \frac {\ln{x}}{\ln{5}} + \frac{\ln{x}}{\ln{8}} = \) \( \displaystyle \ln{x} \cdot \left( \frac{1}{\ln{5}} + \frac{1}{\ln{8}} \right) = \) \( \displaystyle \ln{x} \cdot \frac{ \ln{8} + \ln{5} }{\ln{5} \cdot \ln{8}} = \) \( \displaystyle \frac{\ln{40}}{\ln{5} \cdot \ln{8}} \cdot \ln{x} \)

  8. Solve: \( \displaystyle \sqrt{\log{x}} = \log{\sqrt{x}} \)
    9. Solution

    \( \displaystyle x = {1, 1000} \)

  9. Solve: \( \displaystyle \left( \log_4{x} \right)^2 - 15 \log_4{x} = 16 \)
    10. Solution

    \( \displaystyle x = \left\{ \frac{1}{4}, 4^{16} \right\} \)

  10. Given: \( \displaystyle \log_6{10} = M, \) \( \log_6{7} = P, \) and \( \log_6{4} = Q, \) find \( \log_6{\frac{36}{49}} \) in terms of \(M, P, \) and/or \(Q\)
    11. Solution

    \( \displaystyle 2 - 2P \)

  11. Evaluate: \( \displaystyle \frac{\log_3{\sqrt{243 \sqrt{81 \sqrt[3]{3}}}}}{\log_2{\sqrt[4]{64} + \ln{e^{-10}}}} \)
    12. Solution

    \( \displaystyle - \frac{43}{102} \)

  12. Solve for x: : \( \displaystyle \log_b{x} = 2 - a + \log_b { \left( \frac{a^2b^a}{b^2} \right) } \)
    13. Solution

    \( \displaystyle x = a^2 \)

  13. Find the zero(s) of the graph of \( \displaystyle y = 2\log_2{(x)} - 3 \)
    14. Solution

    \( \displaystyle ( 2 \sqrt{2}, 0) \)

  14. If \( \displaystyle \log_x{a} = 2, \log_x{b} = -3,\) and \( \log_x{c} = 5,\) then evaluate \( \displaystyle \log_x{\frac{ab^3\sqrt{ac}}{c^2}} \)
    15. Solution

    \( \displaystyle -\frac{27}{2} \)

  15. Solve: \( \displaystyle \log{(x + 10)} = \log{x} + \log{10} \)
    16. Solution

    \( \displaystyle x = \frac{10}{9} \)

  16. Solve for \(x\): \( \displaystyle 2\log_b{x} = 2\log_b{(1-a)} + 2\log_b{(1+a)} - \log_b{\left( \frac{1}{a} - a \right)}\)
    17. Solution

    \( \displaystyle x=a \)

  17. Given: \( \displaystyle \log_{\frac{1}{a}}{b}=x. \) Show that \( \displaystyle x = -\log_a{b}\)
    18. Solution

    \( \displaystyle \left( \frac{1}{a} \right)^x = b \implies \) \( \displaystyle \left( a^{-1}\right)^x = b \implies \) \( \displaystyle a^{-x} = b \implies \) \( \displaystyle \log_a{b} = - x \implies \) \( \displaystyle x = -\log_a{b}.\)

  18. Given: \( \displaystyle f(x) = -3\log{(1-4x)} + 1, \) find \( \displaystyle f^{-1}(x) \)
    19. Solution

    \( \displaystyle f^{-1}(x) = \frac{1}{4}\left( 1 - 10^{\frac{1 - x}{3} }\right) \)

  19. Given: \( \displaystyle \log_2{ [ \log_3{(\log_4{x})}]} = \log_3{ [ \log_2{(\log_4{y})}]} = \log_4{ [ \log_3{(\log_2{z})}]} = 0,\) Find \(x + y + z\)
    20. Solution

    \( \displaystyle 88 \)

  20. On the same set of axes, sketch the graphs of \( \displaystyle g(x) = \log_2{x} \) and \( \displaystyle f(x) = 2 \log_2{\left( \frac{1}{2}x - 1 \right)} + 1.\) What is the domain of \( f(x)?\)
    21. Solution

    to check your graph, use your graphing calculator or Desmos. Domain: \( \displaystyle x \gt 2. \)

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  1. The population of the U.S. Virgin Islands has a growth rate of 2.6% per year. In 1990, the populatoin was 512,000. The land area of the Virgin Islands is 3,097,600 square yards. Assuming this growth rate continues and is exponential, when will there be one person for every square yard of land?
    2. Solution

    Approximately 2060

  2. Following the birth of a child, a parent wants to make an initial investment that will grow to $50,000 for the child’s education at age 18. Interest is compounded continuously at 7%. What should be the initial investment?
    3. Solution

    $14,128.70

  3. The decline in the buffalo population was exponential in the 1800’s. In 1800 it was estimated that there were 6,000,000 buffalo roaming the plains. By 1830, this number had declined to 500,000. Assuming the same rate of decline, how many buffalo were there in 1930?
    4. Solution

    126 buffalo

  4. Solve using your calculator: \( \displaystyle \ln{x} = x^2 - 2 \)
    5. Solution

    \( \displaystyle x = \{ 0.13793, 1.56446 \} \)

  5. Solve algebraically. Round your answer to the nearest thousandth: \( \displaystyle 8^{4x - 1} = 205 \)
    6. Solution

    \( \displaystyle x = 0.8900 \)

  6. Solve algebraically. Round your answer to the nearest thousandth: \( \displaystyle 40^{x + 2} = 9^{2x - 4} \)
    7. Solution

    \( \displaystyle x = -2 \)

  7. Solve algebraically. Round your answer to the nearest thousandth: \( \displaystyle e^{3x - 2} = 7^{x - 3} \)
    8. Solution

    \( \displaystyle -3.640 \)

  8. Solve algebraically. Round your answer to the nearest thousandth: \( \displaystyle 3^{x - 5} = 5^{6x+2} \)
    9. Solution

    \( \displaystyle -1.018 \)

  9. Solve algebraically. Round your answer to the nearest thousandth: \( \displaystyle 4e^{1 + 3x} - 9e^{5 - 2x} = 0 \)
    10. Solution

    \( \displaystyle 0.962 \)

  10. Solve algebraically. Round your answer to the nearest thousandth: \( \displaystyle 5(x^2 - 4) = (x^2 - 4) e^{7-x} \)
    11. Solution

    \( \displaystyle 5.391 \)