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How do you solve #log_5 (x+10) + log_5 (x-10) = 3#?

1 Answer

Vinícius Ferraz

Dec 9, 2015

#x in {15}#

Explanation:

#log_5 [(x+10)(x-10)] = log_5 125#

#x^2 - 100 = 125#

#x^2 = 225#

#x = ± 15#

But #x + 10 > 0 and x - 10 > 0#

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