SOLUTION: Log3(1-x)=log3(x+16-x^2)

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Log3(1-x)=log3(x+16-x^2)      Log On


   



Question 1156244: Log3(1-x)=log3(x+16-x^2)
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

We can't tell whether you mean the 3's to be the base of the logarithm
like this:

log%283%2C%281-x%5E%22%22%29%29=log%283%2C%28x%2B16-x%5E2%29%29

or whether you mean the 3's to be multipliers and the logarithms understood
to be common logs with understood base 10, like this:

log%28%283%281-x%5E%22%22%29%5E%22%22%29%29=log%28%283%28x%2B16-x%5E2%29%5E%22%22%29%29 

If it's the first way then we drop the logs and get this:

1-x=x%2B16-x%5E2

If it's the second way then we drop the logs and get this:

3%281-x%5E%22%22%29=3%28x%2B16-x%5E2%29

But if it's that way we divide both sides by 3 and get

1-x=x%2B16-x%5E2

So, luckily here it doesn't matter which you meant but in other
logarithm problems it would make a difference.  So be careful.

1-x=x%2B16-x%5E2

Get 0 on the right and descending order on the left:

x%5E2-2x-15=0

%28x-5%29%28x%2B3%29=0

x-5=0    x%2B3=0
  x=5     x=-3

But we must check in the original equation for extraneous solutions.

If it was the first way, we check to see if 5 is a solution:

log%283%2C%281-5%5E%22%22%29%29=log%283%2C%285%2B16-5%5E2%29%29
log%283%2C%28-4%5E%22%22%29%29=log%283%2C%28-4%5E%22%22%29%29

It looks like it checks but it doesn't because logs cannot be taken of
negative numbers in real number mathematics.  So 5 is extraneous.

If it was the first way, we check to see if -3 is a solution:

log%283%2C%281-%28-3%29%5E%22%22%29%29=log%283%2C%28%28-3%29%2B16-%28-3%29%5E2%29%29
log%283%2C%284%5E%22%22%29%29=log%283%2C%284%5E%22%22%29%29

That checks.  So x = -3 is the only solution.

If you meant the other way, it still would be only x = -3.

Edwin