Question 61958
Solve the equation.
cuberoot2^(2-x) = 2^(x^2)
{{{(2^(2-x))^(1/3)=2^(x^2)}}}
{{{2^((1/3)(2-x))=2^(x^2)}}}  Since the bases are the same, their exponents are equal.
{{{(1/3)(2-x)=x^2}}}
{{{3(1/3)(2-x)=3x^2}}}
{{{2-x=3x^2}}}
{{{2-2-x+x=3x^2+x-2}}}
{{{0=3x^2+x-2}}}  Factor by what ever method you use, I'm using arch,ac,or grouping.
{{{0=3x^2+3x-2x-2}}}
{{{0=(3x^2+3x)+(-2x-2)}}}
{{{0=3x(x+1)-2(x+1)}}}
{{{0=(x+1)(3x-2)}}}  Set each parntheses=0.
x+1=0 and 3x-2=0
x+1-1=0-1 and 3x-2+2=0+2
x=-1 and 3x=2
x=-1 and 3x/3=2/3
x=-1 and x=2/3
:
Check:
for x=-1
cubedroot {{{2^(2-(-1))=2^((-1)^2)}}}
cubed root {{{2^(2+1)=2^1}}}
cubed root {{{2^3=2}}}
{{{2=2}}} x=-1 is valid
for x=2/3
cubed root {{{2^(2-(2/3))=2^((2/3)^2)}}}
cubed root {{{2^(6/3-2/3)=2^(4/9)}}}
cubed root {{{2^(4/3)=2^(4/9)}}}
{{{2^(4/9)=2^(4/9)}}} x=2/3 is also valid.
There's another method for solving this, but you have to use a calculator if you take the log of both sides, so don't freak out if this is not the method you were taught.
Happy Calculating!!!