Question 429393
{{{log(3, ((2x-9)(x+3)))=3}}}
We start by rewriting this in exponential form:
{{{(2x-9)(x+3)=3^3}}}
which simplifies as follows:
{{{2x*x+2x*3+(-9)*x+(-9)*3 = 27}}}
{{{2x^2-3x-27 = 27}}}
This is a quadratic equation so we want one side to be zero. Subtracting 27 from each side we get:
{{{2x^2-3x-54 = 0}}}
Now we factor (or use the Quadratic Formula). This factors fairly easily:
(2x+9)(x-6) = 0
From the Zero Product Property we know that one of the factors must be zero. So:
2x+9 = 0 or x-6 = 0
Solving these we get:
x = -9/2 or x = 6<br>
When solving these equations you must check your answer(s). You must ensure that your answer(s) make all arguments to logarithms positive. Any "solution" that make an argument zero or negative must be rejected!<br>
Use the original equation to check:
{{{log(3, ((2x-9)(x+3)))=3}}}
Checking x = -9/2:
{{{log(3, ((2(-9/2)-9)((-9/2)+3)))=3}}}
which simplifies as follows:
{{{log(3, ((-9-9)((-9/2)+6/2)))=3}}}
{{{log(3, ((-18)(-3/2)))=3}}}
{{{log(3, (27))=3}}}
The argument is positive so there is no reason to reject this solution. (Even though the x was negative, the argument was positive!) The rest of the check is optional and will tell us if we made a mistake. You are welcome to finish the check.<br>
Checking x = 6:
{{{log(3, ((2(6)-9)((6)+3)))=3}}}
which simplifies as follows:
{{{log(3, ((12-9)(9)))=3}}}
{{{log(3, ((3)(9)))=3}}}
{{{log(3, (27))=3}}}
Again, the argument is positive. So again we have no reason to reject this solution.<br>
So there are two solutions to your equation:
x = -9/2 or x = 6