e^x-2 + 4 = 21

I got .8332, is this correct?

2 answers

How did you get that ?

is your equation e^(x-2) + 4 = 21 ?

then
e^(x-2) = 17
ln both sides
ln[e^(x-2)] = ln 17
(x-2) ln e = ln 17
x - 2 = ln 17
x = ln17 + 2 = 4.8332
Oh, i subtracted 2 instead of adding it, thank you.