`3 log_2 (8/15)+ 2 log_2 (25/3)+ log_2 (243/80)` equals

prove that log2 + 16 log(16/15) + 12 log(25/24) + 7 log(81/80) = 1

Prove that "log" 2 + 16 "log" (16)/(15) + 12 "log " (25)/(24) + 7 "log"(81)/(80) = 1 .

det [[log_ (2) 512, log_ (4) 3log_ (3) 8, log_ (3) 9]] xxdet [[log_ (2) 3, log_ (8) 3log_ (3) 4, log_ (3) 4 ]] =

Prove that log2+16log (16/15)+12log(25/24)+7log(81/80)=1

Simplify : 2 log ""( 15)/(8) - log ""( 25)/( 162) +3 log ""( 4)/(9)

(log_(8)17)/(log_(9)23) - (log_(2sqrt(2))17)/(log_(3)23) is equal to