How do you expand #log_bsqrt(57/74)#?

2 Answers
Rachel
Apr 16, 2017

#1/2log_b57-1/2log_b74#

Explanation:

There are certain rules to logratithims. You can find the complete list here, but the one that applies here is the second rule:

#logb(m/n) = logb(m) – logb(n)#

Using this law, we can solve #logbsqrt(57/74)#:
#logb(sqrt(57))/(sqrt(74))#
#logbsqrt(57)-logbsqrt(74)#
We can stop here, but I'm going to keep going and expand it as much as I can
#log_b57^.5-log_b74^.5#
#.5log_b57-.5log_b74#
#1/2log_b57-1/2log_b74#

That's as expanded as it gets! Good work

Monzur R.
Apr 16, 2017

#log_b(sqrt(57/74))=1/2log_b57-1/2log_b74#

Explanation:

#log_b(sqrt(57/74))=log_b((57/74)^(1/2))#

Use the rule #loga^b=bloga#

#log_b((57/74)^(1/2))=1/2log_b(57/74)#

Use the rule #log(a/b)=loga-logb#, and don't forget that the #1/2# belongs to the whole expression.

#1/2log_b(57/74)=1/2log_b57-1/2log_b74#

Related questions
See all questions in Functions with Base b
Impact of this question
7046 views around the world
You can reuse this answer
Creative Commons License