How do you solve #log_5(x - 1) + log_5(x - 2) - log_5(x + 6) = 0#?

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Answer 1 · 2018-06-22T07:14:19

#color(crimson)(x = +- 2sqrt2#

#log_5 (x-1) + log_5 (x-2) - log_5 (x+6) = 0#

#color(blue)(log a + log b = log (ab), log x - log y = log(x/y), " as per log rules"#

# log_5 (((x+1)(x-2)) / (x+6)) = 0#

#((x+1)(x-2)) / (x+6) = 5^0 = 1#

#(x+1) (x-2) = x + 6#

#x^2 -x - 2 = x + 6#

#color(crimson)(x^2 = 8 " or " x = +- 2 sqrt2#