How do you evaluate #log_49 7 + log_27 (1/9) div log_64 (1/32) - log_(3/2) (27/8)#?

1 Answer
Bdub
Mar 22, 2016

#log_49 7 +log_27 (1/9) -: log_64 (1/32) - log_(3/2) (27/8) = -17/10#

Explanation:

#log_49 7 +log_27 (1/9) -: log_64 (1/32) - log_(3/2) (27/8)#
#=(log7/log7^2) +( log 3^-2 /log3^3)-:(log 2^-5 /log 2^6 )-(log (3/2)^3 /log(3/2)) #
#=(log7/(2 log7)) +( (-2 log 3) /(3 log3))-:((-5 log 2) /(6log 2) )-((3log (3/2)) /log(3/2)) #
#=1/2 -2/3-:-5/6 -3#
#=-17/10#

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