How do you solve #e^ { x + 1} - 1= 10#?

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Answer 1 · 2018-05-08T13:06:40

#x=1.399#

By law of logarithms:

#lnx^n# is equivalent to #nln(x)#

Therefore,

#e^(x+1)-1=10#

#e^(x+1)=11#

#lne^(x+1)=ln(11)#

#(x+1)=ln(11)#

#x=ln(11)-1#

#x=1.399#

Double check by substituting the value of #x# back into the original equation and you should get the answer of #10#.