SOLUTION: {{{ln (x-3)=-(x^2)+14}}} is my problem. I know x is approx 3.77, but I need to understand how to get this. If I understand correctly, this is the same as {{{log(e,(x-3))=-(x^2)+1

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: {{{ln (x-3)=-(x^2)+14}}} is my problem. I know x is approx 3.77, but I need to understand how to get this. If I understand correctly, this is the same as {{{log(e,(x-3))=-(x^2)+1      Log On


   

Question 237394: ln+%28x-3%29=-%28x%5E2%29%2B14 is my problem. I know x is approx 3.77, but I need to understand how to get this. If I understand correctly, this is the same as log%28e%2C%28x-3%29%29=-%28x%5E2%29%2B14. I think this can also be e%5E%28-%28x%5E2%29%2B14%29=x-3, but I'm not sure where to go from there. Can you help? I'd really appreciate it!
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
ln+%28x-3%29=-%28x%5E2%29%2B14 is my problem. I know x is approx 3.77, but I need to understand how to get this. If I understand correctly, this is the same as log%28e%2C%28x-3%29%29=-%28x%5E2%29%2B14. I think this can also be e%5E%28-%28x%5E2%29%2B14%29=x-3, but I'm not sure where to go from there. Can you help? I'd really appreciate it!

To 12 decimal places (13 significant figures) the solution
is 3.775482695075

Logarithm functions and exponential functions are not
algebraic functions. These are called transcendental
functions. Whenever variables appear both inside and 
outside of transcendental functions in an equation, 
as in your equation, all methods of algebra fail to
be of any use in finding a solution. The best that
can be done is to solve them by iterative processes, 
and then only an approximate solution can be had.  

Edwin