How do you solve #log_4(x+1) - log_4(x-2) = 3#?

1 Answer
Gió
Dec 6, 2015

I found: #x=2.04762#

Explanation:

You can use the property of logs that tells us:
#logx-logy=log(x/y)#
to get:
#log_4((x+1)/(x-2))=3#
Now we use the definition of log to change it into an exponential:
#(x+1)/(x-2)=4^3#
#(x+1)/(x-2)=64#
#(x+1)=64(x-2)#
#x+1=64x-128#
#63x=129#
#x=129/63=2.04762#

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