How do you solve #log x^2 = (log x)^2#?

Question

959 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-01T21:59:46

#x=100#

Rewrite as:

#2logx=(logx)^2#

#0=(logx)^2-2logx#

#0=logx(logx-2)#

Set each portion equal to #0#.

#=>logx=0#

#x# does not exist.

#=>logx-2=0#

#logx=2#

#log_10x=2#

#10^(log_10x)=10^2#

#x=100#