How to find the value of this equation ?

Question

#log_m 32 * log_2 (1/9) * log_3 m#

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Answer 1 · 2017-09-03T18:37:30

# -10.#

Using the Change of the Base Rule for #log# function, we get,

#log_m32=log_2 32/log_2m=log_2 2^5/log_2m=5/log_2m...(1).#

#log_2(1/9)=log_3(1/9)/log_3 2=log_3(3^-2)/log_3 2=-2/log_3 2...(2).#

#log_3m=log_2 m/log_2 3...............................................................(3).#

#:." The Reqd. Value="(5/log_2m)(-2/log_3 2)(log_2 m/log_2 3),#

#=-10/{(log_3 2)(log_2 3)}.#

#because, log_3 2=log_2 2/log_2 3=1/log_2 3, :. (log_3 2)(log_2 3)=1.#

# rArr " The Reqd. Value="-10.#

Enjoy Maths.!