How do you solve log_5(x - 1) + log_5(x - 2) - log_5(x + 6) = 0log5(x−1)+log5(x−2)−log5(x+6)=0?

1 Answer
sankarankalyanam
Jun 22, 2018

color(crimson)(x = +- 2sqrt2x=±2√2

Explanation:

![https://in.pinterest.com/pin/270849365060467949/?lp=true](useruploads.socratic.org)

log_5 (x-1) + log_5 (x-2) - log_5 (x+6) = 0log5(x−1)+log5(x−2)−log5(x+6)=0

color(blue)(log a + log b = log (ab), log x - log y = log(x/y), " as per log rules"loga+logb=log(ab),logx−logy=log(xy), as per log rules

log_5 (((x+1)(x-2)) / (x+6)) = 0log5((x+1)(x−2)x+6)=0

((x+1)(x-2)) / (x+6) = 5^0 = 1(x+1)(x−2)x+6=50=1

(x+1) (x-2) = x + 6(x+1)(x−2)=x+6

x^2 -x - 2 = x + 6x2−x−2=x+6

color(crimson)(x^2 = 8 " or " x = +- 2 sqrt2x2=8 or x=±2√2

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