How do you solve #log_3(x+4)-log_3x=2#?

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Answer 1 · 2015-12-14T17:32:30

#x = 1/2#

The relevant law of logarithms that will help us here is the following:

#log_z A - log_z B = log_z (A/B)#

where #A# and #B# can be any expression and #z# is the base.

Applying this law to our equation will give us

#log_3 ((x+4)/x) = 2#

From here we can simply raise #3# to both sides of the equation.

#3^(log_3 ((x+4)/x)) = 3^2#

On the left-hand side, the #3# will cancel with the #log_3# leaving us with

#(x+4)/x = 9#

Solving for #x# yields #x = 1/2#.