Question 913163
{{{16^(x) + 4^(x+1) - 3 = 0}}}


{{{(4^2)^x + 4^(x+1) - 3 = 0}}}


{{{4^(2x) + 4^(x+1) - 3 = 0}}}


{{{(4^x)^2 + 4^(x+1) - 3 = 0}}}


{{{(4^x)^2 + (4^x)*(4^(1)) - 3 = 0}}}


{{{(4^x)^2 + (4^x)*(4) - 3 = 0}}}


{{{(4^x)^2 + 4(4^x) - 3 = 0}}}


{{{z^2 +4z - 3 = 0}}} Let {{{z = 4^x}}}


Use the quadratic formula to get {{{z = -2+sqrt(7)}}} or {{{z = -2-sqrt(7)}}}


The steps for the quadratic formula are lengthy and I feel they would clutter up this solution page. Let me know if you need to see the steps to getting those solutions for z.


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The solutions in terms of z are {{{z = -2+sqrt(7)}}} or {{{z = -2-sqrt(7)}}}


We ultimately want solutions in terms of x.


Plug in the first solution for z to get the first solution for x.


{{{z = 4^x}}}


{{{-2+sqrt(7) = 4^x}}}


{{{4^x = -2+sqrt(7)}}}


{{{x = log(4,(-2+sqrt(7)))}}}


So the first solution is {{{x = log(4,(-2+sqrt(7)))}}}


Using a calculator, we get log(-2+sqrt(7))/log(4) = -0.3154747


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Plug in the second solution for z to get the second solution for x.


{{{z = 4^x}}}


{{{-2-sqrt(7) = 4^x}}}


{{{4^x = -2-sqrt(7)}}}


{{{x = log(4,(-2-sqrt(7)))}}}


So the second solution is {{{x = log(4,(-2-sqrt(7)))}}}; however, notice how {{{-2-sqrt(7)=-4.64575}}} is negative. That means the log of this value is undefined. So log(-2-sqrt(7)) = log(-4.64575) = Undefined


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Final Answer: 


The only exact solution is {{{x = log(4,(-2+sqrt(7)))}}}


The only approximate solution is {{{x = -0.3154747}}}



Let me know if you need more help or if you need me to explain a step in more detail.
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Thanks,


Jim