Prove that #log_(2/3)(5/6)# is less that one greater than zero? Also explain your answer?

Question

2347 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-07T03:43:23

Let

#log_(2/3)(5/6)=x#

#=>(2/3)^x=5/6#

Taking log we get

#log_10(2/3)^x=log_10(5/6)#

#=>xlog_10(2/3)=log_10(5/6)#

#=>xlog_10(2/3)=(log_10 5-log_10 6)#

#=>x=(log_10 5-log_10 6)/(log_10 2-log_10 3)#

#=>x=(-(log_10 6-log_10 5))/(-(log_10 3-log_10 2))#

#=>x=log_10 (6/5)/(log_10 (3/ 2))=log_10(1.2)/log_10(1.5)#

obviously # 0< log_10(1.2)/log_10(1.5) < 1#

As we know

#1<1.2 <1.5#

#=>log_10 1 < log_10 1.2 < log_10 1.5#

#=>0< log_10 1.2 < log_10 1.5#

#=>0< (log_10 1.2)/(log_10 1.5) < 1#