How do you find the exact value of #8^(log_8 6-log_8 9)#?

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Answer 1 · 2017-07-26T17:45:51

# 2/3.#

Recall that, #a^x=b....(1). iff x=log_a b..........(2).#

Hence, if we subst. #x# in #(1),# we have, #a^(log_a b)=b.#

Accordingly, #8^(log_8 6)=6, and, 8^(log_8 9)=9.#

Therefore, The Reqd. Value=#8^{(log_8 6)-(log_8 9)},#

#={8^(log_8 6)}/{8^(log_8 9)},#

#=6/9,#

#=2/3.#