Question #a6ad5

Question

870 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-19T18:33:55

#x=2#

#(log2x)/(1-log5)=2 iff log2x=2-2log5 iff#

#log2x=log10^2-log5^2iff log2x=log(10^2/5^2)iff#

#2x=100/25iff2x=4iffx=2#

Answer 2 · 2017-07-19T19:24:02

#x=2#

#"note that "logx-=log_(10)x#

#log2x=2(1-log5)=2-2log5#

#rArrlog2x=log100-log25larr log_(10)100=2#

#rArrlog2x=log(100/25)=log4#

#rArr2x=4rArrx=2#