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December 20, 2014

December 20, 2014

Posted by **Sara** on Wednesday, January 11, 2012 at 7:30pm.

log_7(x-8)- log_7(x-2)=1

- precalculus -
**Damon**, Wednesday, January 11, 2012 at 7:39pmlog7 [(x-8)/(x-2)] = 1

[(x-8)/(x-2)] = 7*1 = 7

x - 8 = 7 x - 14

6 x = 6

x = 1

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