How do you simplify $7^ (log_7 9 - log_7 8)$ ?

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Answer 1 · 2016-03-22T16:41:34

$7^(log_7 9-log_7 8)=9/8$

$7^(log_7 9-log_7 8)$

$=7^(log_7 (9/8)$ -> use property $log_b(x/y)=log_bx-log_by$ for the exponent

$=9/8$ ->use inverse property $b^(log_bx)=x$

How do you simplify 7^ (log_7 9 - log_7 8)7^ (log_7 9 - log_7 8)?