How do you solve the equation $log_9 (x-3)+log_9(3x-2)=log_9 18$ ?

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Answer 1 · 2017-07-06T20:35:48

$x = 4.54647$ to 5 dp

We will need a to use several of the properties of logarithms, namely

$log A + log B -= log AB$
$log A = log B iff A=B$

We have:

$log_9 (x-3)+log_9(3x-2)=log_9 18$

$:. log_9 (x-3)(3x-2)=log_9 18$

$:. (x-3)(3x-2)=18$

$:. 3x^2-11x+6=18$
$:. 3x^2-11x-12=0$

We can solve this quadratic by completing the square or using the quadratic formula, and we get:

$x=11/6+-sqrt(265)/6$
$\ \ = -0.87980, 4.54647$ to 5 dp

Note that we should reject any solution that could have yielded a negative argument to the initial logarithm question

$x-3 \ gt 0 \ \=> x gt 3$
$3x-2 gt 0 => x gt 2/3$

Yielding the only valid solution;

$x = 4.54647$ to 5 dp

How do you solve the equation log_9 (x-3)+log_9(3x-2)=log_9 18log_9 (x-3)+log_9(3x-2)=log_9 18?