How do you condense #20lna-4lnb#?

1 Answer
Aug 23, 2016

Answer:

#ln(a^20/b^4)# or #4ln(a^5/b)#.

Explanation:

#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.