How do you solve #log(x^3)=(logx)^3#?

1 Answer
bp
Sep 11, 2015

x=1, #10^sqrt3, 10^-sqrt3#

Explanation:

#log(x^3)= 3log x#

#3log x= (logx)^3#
#(log x)^3 -3logx =0#

#logx ( (log x)^2-3)#=0

logx=0, #(logx)^2 -3=0#

log x=0 means x=1 and #(logx )^2 -3=0# would mean #log x=+-sqrt3#, or #x=10^sqrt3, 10^-sqrt3#

Related questions
See all questions in Logarithmic Models
Impact of this question
17759 views around the world
You can reuse this answer
Creative Commons License