Logx.log2x=log4x solve equatin?

Question

8578 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-05T05:13:35

#x = 2#

As per the question, we have :

#log x . log 2x# = #log 4x#

#log x . 2x = log 4x#

#log 2x^2 = log 4x#

#2x^2 = 4x#

#2x^2 - 4x = 0#

#x^2 - 2x = 0#

#(x)(x - 2) = 0#

#:.# #x = 0 and 2#

But #x# cannot be #0# as #log 0# is not defined.

#:.# #x = 2#