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

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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#