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How do you solve #e^ { - 4x } + 8= 35#?

1 Answer

Paramecium · Stefan V.

May 18, 2017

You have to use logarithms.

Explanation:

#e^(-4x) + 8 = 35#

#e^(-4x) = 27#

#ln 27 = - 4x#

#(ln 27) / (-4) = x#

#"*"ln# is basically #log_e#

#x# is approximately #-0.8240#.

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