Question 1176692
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We'll apply the rule
{{{(a^b)^c = a^(b*c)}}}
to say that:
{{{9^x = (3^2)^x}}}


{{{9^x = 3^(2x)}}}


{{{9^x = (3^x)^2}}}


The equation
{{{9^x-2(3^x+2)+81=0}}}
then becomes
{{{(3^x)^2-2(3^x+2)+81=0}}}


Let {{{w = 3^x}}}
Replace every copy of {{{3^x}}} with {{{w}}} 


We go from
{{{(3^x)^2-2(3^x+2)+81=0}}}
to
{{{w^2-2(w+2)+81=0}}}


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Let's solve for w


{{{w^2-2(w+2)+81=0}}}


{{{w^2-2w-4+81=0}}}


{{{w^2-2w+77=0}}}


Apply the quadratic formula
{{{w = (-b+sqrt(b^2-4ac))/(2a)}}} or {{{w = (-b-sqrt(b^2-4ac))/(2a)}}}


{{{w = (-(-2)+sqrt((-2)^2-4(1)(77)))/(2(1))}}} or {{{w = (-(-2)-sqrt((-2)^2-4(1)(77)))/(2(1))}}}


{{{w = (4+sqrt(-304))/(2)}}} or {{{w = (4-sqrt(-304))/(2)}}}


We run into a problem. The negative under the square root means we'll have complex numbers. I'll assume your teacher hasn't covered the concept yet (please let me know if otherwise)


So because {{{w^2-2w+77=0}}} has complex solutions, then so does the original equation {{{9^x-2(3^x+2)+81=0}}}



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Now that I think about it further, it's possible that the equation you were given is {{{9^x-2*3^(x+2)+81=0}}} instead of {{{9^x-2*(3^(x)+2)+81=0}}}. Note how the 2 has moved from being in the exponent and then not in the exponent.


If the 2 is in the exponent, then,
{{{9^x-2*3^(x+2)+81=0}}}


{{{(3^x)^2-2*(3^x)*3^(2)+81=0}}}


{{{(3^x)^2-2*(3^x)*9+81=0}}}


{{{w^2-18w+81=0}}}


{{{(w-9)^2=0}}}


{{{w-9 = sqrt(0)}}}


{{{w-9=0}}}


{{{w=9}}}


If w = 9, then, 
{{{w = 3^x}}}


{{{9 = 3^x}}}


{{{3^x = 9}}}


{{{3^x = 3^2}}}


{{{x = 2}}}


So once again, if the equation you were given was {{{9^x-2*3^(x+2)+81=0}}} (note the 2 in the exponent), then it leads to the single solution of {{{x = 2}}}


We can check this by plugging x = 2 back into the original equation to get...
{{{9^x-2*3^(x+2)+81=0}}}


{{{9^2-2*3^(2+2)+81=0}}}


{{{9^2-2*3^4+81=0}}}


{{{81-2*81+81=0}}}


{{{81-162+81=0}}}


{{{-81+81=0}}}


{{{0=0}}}
We get the same thing on both sides, so the answer is confirmed for that equation.
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