How do you solve #e^(x+6) = 4#?

1 Answer
George C.
Jul 17, 2015

Take the natural logarithm of both sides, then subtract #6# from both sides to find:

#x = log_e(4)-6 ~= -4.6137#

Explanation:

Taking #log_e# of both sides we get:

#log_e(4) = log_e(e^(x+6)) = x + 6#

Subtract #6# from both ends to get:

#x = log_e(4) - 6 ~= 1.3863 - 6 = -4.6137#

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