Question 133330
{{{2^(x^2 -5x) = 8}}} Start with the given equation



{{{2^(x^2 -5x) = 2^3}}} Rewrite {{{8}}} as {{{2^3}}}



Since the bases are equal, this means that the exponents are equal



So {{{x^2 -5x=3}}}



{{{x^2 -5x=3}}} Set the exponents equal to one another



{{{x^2-5x-3=0}}}  Subtract 3 from both sides. 



Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{x^2-5*x-3=0}}} ( notice {{{a=1}}}, {{{b=-5}}}, and {{{c=-3}}})





{{{x = (--5 +- sqrt( (-5)^2-4*1*-3 ))/(2*1)}}} Plug in a=1, b=-5, and c=-3




{{{x = (5 +- sqrt( (-5)^2-4*1*-3 ))/(2*1)}}} Negate -5 to get 5




{{{x = (5 +- sqrt( 25-4*1*-3 ))/(2*1)}}} Square -5 to get 25  (note: remember when you square -5, you must square the negative as well. This is because {{{(-5)^2=-5*-5=25}}}.)




{{{x = (5 +- sqrt( 25+12 ))/(2*1)}}} Multiply {{{-4*-3*1}}} to get {{{12}}}




{{{x = (5 +- sqrt( 37 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (5 +- sqrt(37))/2}}} Multiply 2 and 1 to get 2



So now the expression breaks down into two parts


{{{x = (5 + sqrt(37))/2}}} or {{{x = (5 - sqrt(37))/2}}}





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Answer:


So the solutions are 


{{{x = (5 + sqrt(37))/2}}} or {{{x = (5 - sqrt(37))/2}}}