SOLUTION: log3(x+6)+log3(x-6)-log3x=2 solve the logarithmic equation

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Question 747802: log3(x+6)+log3(x-6)-log3x=2
solve the logarithmic equation
Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20855) About Me  (Show Source):
You can put this solution on YOUR website!
log%283%2Cx%2B6%29%2Blog%283%2Cx-6%29-log%283%2Cx%29=2....in base 10 we will have
log%28x%2B6%29%2Flog%283%29%2Blog%28x-6%29%2Flog%283%29-log%28x%29%2Flog%283%29=2


%28log%28x%2B6%29%2Blog%28x-6%29-log%28x%29%29%2Flog%283%29=2
log%28x%2B6%29%2Blog%28x-6%29-log%28x%29=2log%283%29
log%28%28x%2B6%29%28x-6%29%29-log%28x%29=log%283%5E2%29
log%28%28%28x%2B6%29%28x-6%29%29%2Fx%29=log%289%29
%28%28x%2B6%29%28x-6%29%29%2Fx=9
%28x%2B6%29%28x-6%29=9x

x%5E2-6x%2B6x-36=9x
x%5E2-36=9x
x%5E2-9x-36=0..write -9x as -12x%2B3x
x%5E2-12x%2B3x-36=0....group
%28x%5E2-12x%29%2B%283x-36%29=0
x%28x-12%29%2B3%28x-12%29=0
%28x%2B3%29%28x-12%29=0
solutions:
if x%2B3=0 => x=-3
if x-12=0 => x=12

recall:
logarithm of negative number is undefined
so, your solution is only highlight%28x=12%29





Answer by MathTherapy(10801) About Me  (Show Source):
You can put this solution on YOUR website!
log3(x+6)+log3(x-6)-log3x=2
solve the logarithmic equation
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It's heart-wrenching to see how the other person (@MATHLOVER) makes this problem so long and time-consuming! Why??
In my opinion, she should learn how to solve problems much more efficiently.
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log+%283%2C+%28x+%2B+6%29%29+%2B+log+%283%2C+%28x+-+6%29%29+-+log+%283%2C+%28x%29%29+=+2
This log equation starts with 3 log arguments: a1 (x + 6), a2 (x - 6), and a3 (x), the smallest being log argument
a2, or x - 6. Argument "a2" "tells" one that x - 6 MUST be > 0. So, x - 6 > 0_____x > 6 
We now have: log+%283%2C+%28x+%2B+6%29%29+%2B+log+%283%2C+%28x+-+6%29%29+-+log+%283%2C+%28x%29%29+=+2, with x > 6           
                       log+%283%2C+%28%28%28x+%2B+6%29%28x+-+6%29%29%2F%28x%29%29%29+=+2 ---- Applying log+%28b%2C+%28a1%29%29+%2B+log+%28b%2C+%28a2%29%29+-+log+%28b%2C+%28a3%29%29 = log+%28b%2C+%28%28a1%2Aa2%29%29%2Fa3%29

                               %28x+%2B+6%29%28x+-+6%29%2Fx+=+3%5E2 --- Applying c+=+b%5Ed, when: log+%28b%2C+%28c%29%29+=+d
                               %28x+%2B+6%29%28x+-+6%29%2Fx+=+9
                                  %28x%5E2+-+36%29%2Fx+=+9
                                    x%5E2+-+36+=+9x --- Cross-multiplying
                                x%5E2+-+9x+-+36+=+0
                           (x - 12)(x + 3) = 0
                            x - 12 = 0       OR      x + 3 = 0
                                 x = 12      OR          x = - 3 

As stated above, x MUST be > 6, so ONLY x = 12 is ACCEPTABLE. This makes x = - 3, an EXTRANEOUS solution to this
equation, and is therefore IGNORED/REJECTED.