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How do you solve #-1-8ln(-8b-5)=7#?

1 Answer

Apr 30, 2018

#b = 1/8(1/e + 5)#

Explanation:

#-1-8ln(-8b-5)=7#

# ln(-8b-5)= -1#

#"Note" : log_b a = c => a = b^c#

# -8b-5 = e^-1#

#b = 1/8(1/e + 5)#

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