How do i write this as a single logarithim using the properties of logarithims? #log_5 9+log_5 8-1#

1 Answer
MeneerNask
Mar 10, 2018

#log_x xhArr1# and vice versa.

Explanation:

So we now have:
#=log_5 9+log_5 8 -log_5 5#

We use the properties:
#log(axxb)hArrloga + logb#
#log(a/b)hArrloga - logb#

So:
#=log_5 (9xx8) -log_5 5#

#=log_5 ((9xx8)/5)=log_5 (72/5)=log_5 14.4#

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