Question 113823
{{{2^(x^2+x)=64}}} Start with the given equation



{{{2^(x^2+x)=2^6}}} Rewrite {{{64}}} as {{{2^6}}}



Since the bases are equal, the exponents are equal. So {{{x^2+x=6}}}


{{{x^2+x=6}}} Now let's solve for x



{{{x^2+x-6=0}}} Subtract x from both sides




{{{(x+3)(x-2)=0}}} Factor the left side (note: if you need help with factoring, check out this 
<a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:


{{{x+3=0}}} or {{{x-2=0}}}


{{{x=-3}}} or {{{x=2}}}  Now solve for x in each case



So our solutions are {{{x=-3}}} or {{{x=2}}}



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Check:

Let's check the solution {{{x=-3}}}


{{{2^(x^2+x)=64}}} Start with the given equation



{{{2^((-3)^2+(-3))=64}}} Plug in {{{x=-3}}}



{{{2^(9+(-3))=64}}} Square -3 to get 9



{{{2^6=64}}} Add


{{{64=64}}} Raise 2 to the sixth power to get 64. Since both sides of the equation are equal, this solution is verified.


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Let's check the solution {{{x=2}}}


{{{2^(x^2+x)=64}}} Start with the given equation



{{{2^((2)^2+(2))=64}}} Plug in {{{x=2}}}



{{{2^(9+(2))=64}}} Square 2 to get 4



{{{2^6=64}}} Add


{{{64=64}}} Raise 2 to the sixth power to get 64. Since both sides of the equation are equal, this solution is verified.



So this verifies our solutions {{{x=-3}}} and {{{x=2}}}