Question 1156244
<pre>

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

{{{log(3,(1-x^""))=log(3,(x+16-x^2))}}}

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((3(1-x^"")^""))=log((3(x+16-x^2)^""))}}} 

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

{{{1-x=x+16-x^2}}}

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

{{{3(1-x^"")=3(x+16-x^2)}}}

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

{{{1-x=x+16-x^2}}}

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+16-x^2}}}

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

{{{x^2-2x-15=0}}}

{{{(x-5)(x+3)=0}}}

{{{x-5=0}}}    {{{x+3=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(3,(1-5^""))=log(3,(5+16-5^2))}}}
{{{log(3,(-4^""))=log(3,(-4^""))}}}

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(3,(1-(-3)^""))=log(3,((-3)+16-(-3)^2))}}}
{{{log(3,(4^""))=log(3,(4^""))}}}

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

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

Edwin</pre>