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= $8^{(log_8 6)-(log_8 9)},$

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

$=6/9,$

$=2/3.$

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