How do you solve #log_2 (x + 3) - log_2 (x – 4) = log_2 8#?

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Answer 1 · 2016-03-25T02:05:05

#x=5#

Since all the logs have the same base you can take advantage of the property: #log_n(a/b)= log_n(a) - log_n(b) #

Rewrite the equation:

#log_2((x+3)/(x-4)) = log_2(8)#

Rewrite according to: #log_b(x)=y, b^y = x#

#2^(log_2(8)) = (x+3)/(x-4)#

Note that #a^(log_a(b)) = b#, so you can rewrite as:

#8 = (x+3)/(x-4)#
Solve for x:
#8(x-4) = (x+3)#
#8x-32 = x+3#
#7x-32=3#
#7x=35#
#x = 5#