How do you solve $log_3(x-6)=log_9(2x)$ ?

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Answer 1 · 2016-09-30T16:13:34

$x = 7+ sqrt(13)$

$log_3(x- 6) - log_9(2x) = 0$

$log(x - 6)/log3 - log(2x)/log9 = 0$

$log(x- 6)/log3 - log(2x)/(2log3) = 0$

Put on a common denominator:

$(2log(x - 6))/(2log3) - log(2x)/(2log3) = 0$

$log_9(x - 6)^2- log_9(2x) = 0$

$log_9((x^2 - 12x + 36)/(2x)) = 0$

$(x^2 - 12x + 36)/(2x) = 9^0$

$(x^2 - 12x + 36)/(2x) = 1$

$x^2 - 12x + 36 = 2x$

$x^2 - 14x + 36 = 0$

$1(x^2 - 14x + 49 - 49) = -36$

$1(x - 7)^2 - 49 = -36$

$1(x - 7)^2 = 13$

$(x - 7) = +-sqrt(13)$

$x = 7 +- sqrt(13)$

However, the $x = 7 - sqrt(13)$ is extraneous, because it renders the initial equation undefined.

Hopefully this helps!

How do you solve log_3(x-6)=log_9(2x)log_3(x-6)=log_9(2x)?