I guess you're supposed to write the expression on the right in terms of [tex]A,B,C[/tex].
18. Notice that [tex]729=9^3[/tex], so
[tex]\log_8729=\log_89^3=3\log_89=3B[/tex]
20. Notice that [tex]\dfrac{14}3=\dfrac{42}9=\dfrac{6\times7}9[/tex], so
[tex]\log_8\dfrac{14}3=\log_8\dfrac{6\times7}9=\log_86+\log_87-\log_89=A+C-B[/tex]
