Question #a59cd

1 Answer
Sanjay VN
Jan 15, 2018

is 0

Explanation:

so
#log10^x = log5^(2x)#
#log10^x - log5^(2x)=0#
which is
#log(10^x/5^(2x))=0#
#log((2*5)^x/(5^x*5^x))=0#
#log(2^x/5^x)=0#
hence
#(2/5)^x = 1#
so
#2^x = 5^x#
which is never possible for any value other than
#x=0#
therefore solution is #x=0#
hope u find it helpful :)

Related questions
Impact of this question
836 views around the world
You can reuse this answer
Creative Commons License