Question 753630
Consider the function {{{g(y)=(5+2sqrt(2))^y+(5-2sqrt(2))^y}}}
 
{{{g(1)=(5+2sqrt(2))^1+(5-2sqrt(2))^1=5+2sqrt(2)+5-2sqrt(2)=10}}}
 
{{{g(y)=(5+2sqrt2)^y+(5-2sqrt2)^y}}} is the sum of two exponential functions,
with positive bases, {{{0<5-2sqrt2<5+2sqrt2}}}.
Those two exponmential functions, and {{{g(y)}}}, increase with {{{y}}} throughout their real numbers domain, from -infinity to infinity.
 
So for {{{y<1}}} --> {{{g(y)<10}}}
and for {{{y>1}}} --> {{{g(y)>10}}}
The only value of {{{y}}} that makes {{{g(y)=10}}} is {{{y=1}}}
 
Then, the only solutions for {{{(5+2sqrt(2))^(x^2-x)+(5-2sqrt(2))^(x^2-x)=10}}}
will be the solutions of {{{x^2-x=1}}}
 
{{{x^2-x=1}}} --> {{{x^2-x-1=0}}}
Applying the quadratic formula, we find
{{{x=(-1 +- sqrt(1^2-4(1)(-1)))/2/1}}} --> {{{highlight(x=(1 +- sqrt(5))/2)}}}