Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
How do you condense ##20lna-4lnb##?
##ln(a^20/b^4)## or ##4ln(a^5/b)##.
##20lna=lna^20## by the rule ##bloga=loga^b##.
Similarly, ##4lnb=lnb^4## by the same rule.
So now we have ##20lna-4lnb=lna^20-lnb^4##.
Now, using the rule ##loga-logb=log(a/b)##, we can rewrite this as ##ln(a^20/b^4)##.
If you like, you can write this as ##ln((a^5/b)^4)##, which can be rewritten as ##4ln(a^5/b)##.
You can plug numbers in to these rules to make sure they work. For example, ##log_2 32-log_2 8=5-3=2##. If you use the division rule, you get ##log_2 32-log_2 8=log_2(32/8)=log_2(4)=2##.
More generally, we can show this rule and the others to be true. I will show that ##loga+logb=log(a*b)##.
Start with ##y=lna+lnb##.
Exponentiate both sides to get ##e^y=e^(lna+lnb)##.
Through rules of exponents, the right side can be rewritten to get ##e^y=e^lna*e^lnb##.
Since ##e^lnx=x## (if you're confused why, say something) we get ##e^y=a*b##.
To finish, take the natural log of both sides to get ##y=ln(a*b)##.
Similar things can be done to prove the other rules true.