How do you solve this ?

Question

#log_m 32 * log_(2) (1/9) * log_3 m#

Please do not use natural logarithm
Just use any base logarithms

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Answer 1 · 2017-09-04T02:39:10

#log_m32*log_2(1/9)*log_3m#

#=log_m2^5*log_2(1/9)*log_3m#

#=5log_m2*log_2(1/9)*log_3m#

#=5log_m(1/9)*log_3m#

#=5log_3(1/9)#

#=5log_3 3^-2#

#=5(-2)log_3 3=-10*1=-10#

Note:

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