How do you solve $log_3 x = log_3(x + 2) = log_3 2 + log_3 12$ ?

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Answer 1 · 2015-09-06T14:33:17

It is not obvious what the equation should be.

It seem likely that the equation should be

$log_3 x + log_3(x + 2) = log_3 2 + log_3 12$

$log_3(x(x+2)) = log_3 (2*12)$ $" "$ (property of logarithms)

$(x(x+2)) = (2*12)$ $" "$ (logarithms are one-to-one)

$x^2+2x = 24$

$x^2+2x-24=0$

$(x+6)(x-4) = 0$

$x=-6$ is an extraneous solution. $log_3(-6)$ is not defined (in the real numbers)

$x=4$ is the only solution.

How do you solve log_3 x = log_3(x + 2) = log_3 2 + log_3 12 log_3 x = log_3(x + 2) = log_3 2 + log_3 12?