How do you solve #e^(x+1) = 30#?

1 Answer
mason m
Dec 30, 2015

#x=ln(30)-1approx2.4012#

Explanation:

In order to undo the exponentiation, take the natural logarithm of both sides. (The natural logarithm, denoted #ln#, is equivalent to #log_e#.) Since logarithms are the inverse of exponential functions, taking the logarithm will isolate the #x# term.

#e^(x+1)=30#

#ln(e^(x+1))=ln(30)#

#x+1=ln(30)#

#x=ln(30)-1approx2.4012#

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