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Correct.

Correct. :)
3. ### 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. ### 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)}

6. ### 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)}

8. ### Practice In Terms of A, B, and/or C 

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

Correct.

Correct.
11. ### Practice In Terms of A, B, and/or C 

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

Correct.
13. ### Practice Sum & Difference of Logs 7

I feel like I just saw this?
14. ### Practice Sum & Difference of Logs 6

You didn't distribute the negative correctly.

Correct.

Correct.

Correct.
18. ### 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. ### Practice Solve Equations

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