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

1 Answer
May 15, 2016

Answer:

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#