How do you evaluate # Log_3 9 + log_3 36- log_3 4#?

1 Answer
P dilip_k
May 15, 2016

4

Explanation:

Using formula#" "log_am +log_an =log_a(mxxn)#

# log_3 9 + log_3 36- log_3 4#

# =log_3 (9xx36)- log_3 4 " "#

# =log_3( (9xxcancel36^9)/cancel 4) " "# Using formula#" "log_am +log_an =log_a(mxxn)#

# =log_3 (3^4)=4log_3 3=4xx1=4#

Using formula#" "log_aa =1#

Related questions
See all questions in Functions with Base b
Impact of this question
2765 views around the world
You can reuse this answer
Creative Commons License