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  1. MarkFL

    Practice Change of Base Formula 5

    Correct.
  2. MarkFL

    Practice Change of Base Formula 4

    Correct. :)
  3. MarkFL

    Practice Change of Base Formula 3

    You have incorrectly distributed a negative sign again. \log_b\left(\frac{5}{9}\right)=\log_b(5)-2\log_b(3)=C-2B
  4. MarkFL

    Practice Change of Base Formula 2

    Yes, one implication of the change of base theorem is that: \log_a(b)=\frac{1}{\log_b(a)}
  5. MarkFL

    Practice Change of Base Formula 1

    \log_{3}(10b)=\frac{\log_b(10b)}{\log_b(3)}=\frac{\log_b(2)+\log_b(5)+\log_b(b)}{\log_b(3)}=\frac{A+C+1}{B}\quad\checkmark
  6. MarkFL

    Practice Express As Natural Logs

    Using the change of base formula: 1.) \log_2(10)=\frac{\ln(10)}{\ln(2)} 2.) \log_{e^2}(2)=\frac{\ln(2)}{\ln(e^2)}=\frac{\ln(2)}{2} 3.) \log_{10}(\log_{10}(x))=\frac{\ln(\log_{10}(x))}{\ln(10)}=\frac{\ln\left(\dfrac{\ln(x)}{\ln(10)}\right)}{\ln(10)}
  7. MarkFL

    Practice Sum & Difference of Log 4

    Can you link to the threads?
  8. MarkFL

    Practice In Terms of A, B, and/or C [4]

    You are consistently distributing incorrectly. Recall: -(a+b+c)=-a-b-c
  9. MarkFL

    Practice In Terms of A, B, and/or C [3]

    Correct.
  10. MarkFL

    Practice In Terms of A, B, and/or C [2]

    Correct.
  11. MarkFL

    Practice In Terms of A, B, and/or C [1]

    Correct. These images are huge and take a long time for me to download...can you not scale them down?
  12. MarkFL

    Practice Sum & Difference of Logs 8

    Correct.
  13. MarkFL

    Practice Sum & Difference of Logs 7

    I feel like I just saw this?
  14. MarkFL

    Practice Sum & Difference of Logs 6

    You didn't distribute the negative correctly.
  15. MarkFL

    Practice Sum & Difference of Logs 5

    Correct.
  16. MarkFL

    Practice Sum & Difference of Log 4

    Correct.
  17. MarkFL

    Practice Sum & Difference of Logs 3

    Correct.
  18. MarkFL

    Practice Express As Base 10 Logs

    1.) Correct. We could posit that: \log_b(a)=\frac{\log_a(a)}{\log_a(b)}=\frac{1}{\log_a(b)} 2.) Using the above identity, we may then write: \ln(10)=\frac{1}{\log_{10}(e)} 3.) Correct.
  19. MarkFL

    Practice Solve Equations

    Those are correct, but keep in mind: \ln(x)\equiv\log_e(x) And: \log_b(27)=3\log_b(3)
  20. MarkFL

    Practice Express In Terms of t and u [2]

    1.) \ln\left(\frac{ex}{y}\right)-\ln\left(\frac{y}{ex}\right)=2\ln\left(\frac{ex}{y}\right)=2(1+\ln(x)-\ln(y))=2(1+t-u)\quad\checkmark 2.)...