What is the value of #((log_2 11)(log_3 12)(log_5 13))/((log_5 11)(log_8 12) (log_9 13)#?

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Answer 1 · 2016-09-16T04:29:25

6

Use# log_b a = log a/log b and log m^n=n log m#..

Now, the given expression is

#((log 11/log 2)(log 12/log 3)(log 13/log 5))/((log 11/log 5)(log12/log 8)(log 13/log 9))#

#=(log 8 log 9)/(log 2 log 3)#

#=(log 2^3 log 3^2)/(log 2 log 3)#

#=((3 log 2)(2 log 3))/(log 2 log 3)#

#=6#.