How do you simplify #log_5 (17/8) log_5 (51/16)#?

Question

1502 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-04T07:15:59

#log_5(17/8)log_5(51/16)=1.18862#

#log_5(17/8)log_5(51/16)#

= #log_5(17/8xx51/16)#

= #log_5((17xx17xx3)/(8xx8xx2))#

= #2log_5(17/8)+log_5(3/2)# as #loga^nb=nloga+logb#

Now as #log_5u=logu/log5#, this can be written as

#(2log(17/8)+log(3/2))/log5#

= #(2log(2.125)+log(1.5))/log5#

= #(2xx0.32736+0.17609)/(0.69897)#

= #1.18862#