SOLUTION: find x, if; {{{ (5+2sqrt2)^(x^2-x) + (5-2sqrt2)^(x^2-x) =10 }}}

Algebra ->  Equations -> SOLUTION: find x, if; {{{ (5+2sqrt2)^(x^2-x) + (5-2sqrt2)^(x^2-x) =10 }}}      Log On


   

Question 753630: find x, if;
+%285%2B2sqrt2%29%5E%28x%5E2-x%29+%2B+%285-2sqrt2%29%5E%28x%5E2-x%29+=10++
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Consider the function g%28y%29=%285%2B2sqrt%282%29%29%5Ey%2B%285-2sqrt%282%29%29%5Ey



g%28y%29=%285%2B2sqrt2%29%5Ey%2B%285-2sqrt2%29%5Ey is the sum of two exponential functions,
with positive bases, 0%3C5-2sqrt2%3C5%2B2sqrt2.
Those two exponmential functions, and g%28y%29, increase with y throughout their real numbers domain, from -infinity to infinity.

So for y%3C1 --> g%28y%29%3C10
and for y%3E1 --> g%28y%29%3E10
The only value of y that makes g%28y%29=10 is y=1

Then, the only solutions for %285%2B2sqrt%282%29%29%5E%28x%5E2-x%29%2B%285-2sqrt%282%29%29%5E%28x%5E2-x%29=10
will be the solutions of x%5E2-x=1

x%5E2-x=1 --> x%5E2-x-1=0
Applying the quadratic formula, we find
x=%28-1+%2B-+sqrt%281%5E2-4%281%29%28-1%29%29%29%2F2%2F1 --> highlight%28x=%281+%2B-+sqrt%285%29%29%2F2%29