Question 133343
{{{2^(2x^2-3x)= 2^(x^2 - 2x + 12)}}} Start with the given equation



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


{{{2x^2-3x=x^2 - 2x + 12}}} Set the exponents equal to one another



{{{0=x^2-2x+12-2x^2+3x}}}  Subtract {{{2x^2}}} from both sides.  Add 3x to both sides. 



{{{0=-x^2+x+12}}}  Combine like terms




{{{0=-(x-4)(x+3)}}} Factor the right side 



Now set each factor equal to zero:

{{{x-4=0}}} or  {{{x+3=0}}} 


{{{x=4}}} or  {{{x=-3}}}    Now solve for x in each case



So our answers are


 {{{x=4}}} or  {{{x=-3}}}