Question 1206117
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{{{log(3,(x))+log(3,(x+5)) = 9}}}


{{{log(3,(x(x+5))) = 9}}} use log rule that log(A)+log(B) = log(A*B)


{{{x(x+5) = 3^9}}} convert from log form to exponent form


{{{x(x+5) = 19683}}}


{{{x^2+5x = 19683}}}


{{{x^2+5x-19683 = 0}}}


Apply the quadratic formula where a = 1, b = 5, c = -19683
{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-5+-sqrt((5)^2-4(1)(-19683)))/(2(1))}}}


{{{x = (-5+-sqrt(25 + 78732))/(2)}}}


{{{x = (-5+-sqrt(78757))/(2)}}}


{{{x = (-5+sqrt(78757))/(2)}}} or {{{x = (-5-sqrt(78757))/(2)}}}


{{{x = 137.818388}}} or {{{x = -142.818388}}} both of which are approximate


Since the domain of log(x) is x > 0, it rules out {{{x = (-5-sqrt(78757))/(2) = -142.818388}}} as a possible solution.


Therefore, we go with {{{x = (-5+sqrt(78757))/(2) = 137.818388}}} as the only possible solution.


I used WolframAlpha to confirm the answer is correct.
<a href="https://www.wolframalpha.com/input?i=log_%283%29x%2Blog_%283%29%28x%2B5%29%3D9">https://www.wolframalpha.com/input?i=log_%283%29x%2Blog_%283%29%28x%2B5%29%3D9</a>


Another way to confirm the answer is to let
f(x) = log(x)/log(3) + log(x+5)/log(3)
Note the use of the change of base formula.
Then you should find that f(137.818388) = 9 approximately.
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