How do you solve this ?

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

Please do not use natural logarithm
Just use any base logarithms

1 Answer
Sep 4, 2017

Explanation:

#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:

Formulae used

  • #log_ab^n=nlog_ab#

  • #log_ab*log_ca=log_cb#