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How do you solve #log _2 (x+2) - log _2 (x-5) = 3#?

1 Answer

A. S. Adikesavan

May 20, 2016

6

Explanation:

Use, #log_bm-log_bn=log_b(m/n)# and, if #log_b a=c# then #a=b^c#

#log_2 (x+2)-log_2 (x-5)#

#=log_2((x+2)/(x-5))=3#. Inversely,

(x+2)/(x-5)= 2^3=8#. Solving,.

x=6..

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