How do you solve #log_6 u–log_6 9=2#?

1 Answer
Gió
Feb 4, 2015

The result should be #u=324#

You start using a property of logarithms where you have:
#log_a(x)-log_a(y)=log_a(x/y)#

and the basic definition of logarithm: #log_a(x)=b# #-># #x=a^b#

And using these in your example:

#log_6(u/9)=2# and so:
#u/9=6^2#
#u=9*6^2#
#u=324#

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