How do you solve #Ln e^x = 5#?

Question

2983 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-16T19:36:48

Real solution: #x=5#

Complex solutions: #x=5+2kpi i " "# for any #k in ZZ#

The only Real value of #x# for which #ln e^x = 5# is #5#.

The function #x |-> e^x# is one to one from #(-oo, oo)# onto #(0, oo)#.

The function #x |-> ln x# is its inverse from #(0, oo)# onto #(-oo, oo)#.

So for any #x in (-oo, oo)# we have #ln e^x = x#.

#color(white)()#
If we consider Complex solutions, note that #e^(2pii) = 1#

Hence we find solutions:

#x = 5 + 2kpii " "# for any integer #k#

since:

#ln e^(5+2kpii) = ln (e^5 e^(2kpii)) = ln(e^5 * (e^(2pii))^k) = ln (e^5*1^k) = ln e^5 = 5#