How do you solve #e^x-9=19#?

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Answer 1 · 2017-08-30T08:22:18

Real solution:

#x = ln 28#

Complex solutions:

#x = ln 28 + 2npii" "n in ZZ#

Given:

#e^x-9 = 19#

Add #9# to both sides to get:

#e^x = 28#

Take the natural log of both sides to get:

#x = ln(28)#

More generally note that:

#e^(x+2npii) = e^x*e^(2npii) = e^x" "# for any integer #n#

So if we include complex solutions, we have:

#x = ln 28+2npii" "n in ZZ#