How do you condense #[1 - 5log_3x]/2 #?

1 Answer
Jim H
Aug 29, 2015

#log_3 (3/x^5)^(1/2) " or " log_3 sqrt(3/x^5)#

Explanation:

I assume that "condense" means "write as a single logarithm of a single expression". (I assume that because that's what I mean when I tell students to condense such an expression.)

#[1 - 5log_3x]/2 = 1/2(1-5log_3 x)#

# = 1/2 (1-log_3 x^5)#

# = 1/2 (log_3 3 - log_3 x^5)#

# = 1/2 log_3 (3/x^5)#

# = log_3 (3/x^5)^(1/2) " or " log_3 sqrt(3/x^5)#

Related questions
See all questions in Functions with Base b
Impact of this question
1538 views around the world
You can reuse this answer
Creative Commons License