SOLUTION: solve the logarithm 2log3z-log3(z-2)=2

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Question 174880: solve the logarithm

2log3z-log3(z-2)=2
Found 2 solutions by jim_thompson5910, josmiceli:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
2%2Alog%283%2C%28z%29%29-log%283%2C%28z-2%29%29=2 Start with the given equation.


Take note that z%3E0 (for the first log) and z-2%3E0---> z%3E2 (for the second log).


So the domain is z%3E2 (this interval works for BOTH logs)


log%283%2C%28z%5E2%29%29-log%283%2C%28z-2%29%29=2 Rewrite the first log using the identity y%2Alog%28b%2C%28x%29%29=log%28b%2C%28x%5Ey%29%29


log%283%2C%28%28z%5E2%29%2F%28z-2%29%29%29=2 Combine the logs using the identity log%28b%2C%28A%29%29-log%28b%2C%28B%29%29=log%28b%2C%28A%2FB%29%29


3%5E2=%28z%5E2%29%2F%28z-2%29 Rewrite the equation using the property: log%28b%2C%28x%29%29=y ====> b%5Ey=x


9=%28z%5E2%29%2F%28z-2%29 Square 3 to get 9


9%28z-2%29=z%5E2 Multiply both sides by z-2.


9z-18=z%5E2 Distribute.


-z%5E2%2B9z-18=0 Subtract z%5E2 from both sides.


Notice we have a quadratic equation in the form of az%5E2%2Bbz%2Bc where a=-1, b=9, and c=-18


Let's use the quadratic formula to solve for z


z+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


z+=+%28-%289%29+%2B-+sqrt%28+%289%29%5E2-4%28-1%29%28-18%29+%29%29%2F%282%28-1%29%29 Plug in a=-1, b=9, and c=-18


z+=+%28-9+%2B-+sqrt%28+81-4%28-1%29%28-18%29+%29%29%2F%282%28-1%29%29 Square 9 to get 81.


z+=+%28-9+%2B-+sqrt%28+81-72+%29%29%2F%282%28-1%29%29 Multiply 4%28-1%29%28-18%29 to get 72


z+=+%28-9+%2B-+sqrt%28+9+%29%29%2F%282%28-1%29%29 Subtract 72 from 81 to get 9


z+=+%28-9+%2B-+sqrt%28+9+%29%29%2F%28-2%29 Multiply 2 and -1 to get -2.


z+=+%28-9+%2B-+3%29%2F%28-2%29 Take the square root of 9 to get 3.


z+=+%28-9+%2B+3%29%2F%28-2%29 or z+=+%28-9+-+3%29%2F%28-2%29 Break up the expression.


z+=+%28-6%29%2F%28-2%29 or z+=++%28-12%29%2F%28-2%29 Combine like terms.


z+=+3 or z+=+6 Simplify.



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Answer:


So the solutions are z+=+3 or z+=+6


Take note that both of these solutions are within the domain z%3E2. So they are valid solutions.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
2log%283%2Cz%29+-+log%283%2C%28z-2%29%29+=+2
log%283%2Cz%5E2%29+-+log%283%2C%28z-2%29%29+=+2
log%283%2C+z%5E2%2F%28z-2%29%29+=+2
3%5E2+=+z%5E2%2F%28z-2%29
9%2A%28z+-+2%29+=+z%5E2
z%5E2+-+9z+%2B+18+=+0
%28z+-+3%29%28z+-+6%29+=+0
z+=+3
z+=+6
These solutions check if I plug them back in