How do you use the properties of logarithms to expand #log_3(sqrt(a-1)/9)#?

Question

1375 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-08T02:14:52

#1/2log_3 (a - 1) - 2#

Use the following laws of logarithms to solve this problem

•#loga^n = nloga#
•#log(a/b) = loga - logb#
#•log_an =logn/loga#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#=log_3sqrt(a - 1) - log_3 9#

#=log_3(a - 1)^(1/2) - log_3 9#

#= 1/2log_3(a - 1) - log9/log3#

#= 1/2log_3(a - 1) - log3^2/log3#

#=1/2log_3(a - 1) - (2log3)/log3#

#=1/2log_3(a - 1) - 2#

Hopefully this helps!