How do you solve #4.12=e^(-2x)#?

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Answer 1 · 2016-07-11T13:35:31

#x = -0.71# to 2dp

#4.12 = e^(-2x)#

Taking natural log of both sides

#ln(4.12) = lne^(-2x)#

Can rewrite using rules of logs as

#ln(4.12) = -2xlne#

ln and exp are inverse operations so cancel out:

#ln(4.12) = -2x implies x = -1/2ln(4.12)#

#therefore x = -0.71# to 2dp