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How do you solve #log x + log (x-3)=1#?

1 Answer

Vinícius Ferraz

Dec 6, 2015

#b = 10 Rightarrow x = (3 + 7)/2 = 5#

Explanation:

#log_b [x(x-3)] = log_b b#

#x(x-3) = b#

#x^2 -3x - b = 0#

#x = (3 ± sqrt{9 + 4b})/2#

Consider #x > 0# and #x - 3 > 0#

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