How do you solve #Logx+lnx=1#?

Question

3040 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-28T13:40:04

#x = 2.00814#

#log_(10)x = log_e x/(log_e 10)#

so

# log_e x/(log_e 10)+log_e x = 1#

then

#log_e x(1/log_e 10+1)=1#

#log_e x = log_e 10/(1+log_e 10)#

#x = e^( log_e 10/(1+log_e 10)) = 10^(1/(1+log_e 10)) = 2.00814#