How do you simplify #7^ (log _(7)9 - log_(7)8)#?

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Answer 1 · 2016-05-10T11:55:34

#9/8#

Say that this equation equals #x#, so

#7^(log_7 9-log_7 8)=x#

Now take the #log_7# of both sides to get rid of the powers,

#log_7 9-log_7 8=log_7 x#.

We know that #loga-logb=log(a/b)#, so

#log_7 (9/8)=log_7 x#

Raise both sides by the base of #7# to remove logarithms, and find the answer

#9/8=x#