Question 143984
{{{4^(2x)+4^(x+1)-6=0}}} Start with the given equation



{{{4^(2x)+4^(x)*4^(1)-6=0}}} Rewrite {{{4^(x+1)}}} as {{{4^(x)*4^(1)}}} using the identity {{{x^(y+z)=x^(y)*x^(z)}}}



{{{4^(2x)+4^(x)*4-6=0}}} Evaluate {{{4^1}}} to get 4



{{{4^(2x)+4*4^(x)-6=0}}} Rearrange the terms



{{{(4^(x))^2+4*4^(x)-6=0}}} Rewrite {{{4^(2x)}}} as {{{(4^(x))^2}}} using the identity {{{x^(y*z)=(x^(y))^z}}}



Let {{{u=4^x}}}



{{{u^2+4*u-6=0}}} Plug in {{{u=4^x}}}. In other words, replace each instance of {{{4^x}}} with "u"



Let's use the quadratic formula to solve for "u":



Starting with the general quadratic


{{{au^2+bu+c=0}}}


the general solution using the quadratic equation is:


{{{u = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{u^2+4*u-6=0}}} ( notice {{{a=1}}}, {{{b=4}}}, and {{{c=-6}}})





{{{u = (-4 +- sqrt( (4)^2-4*1*-6 ))/(2*1)}}} Plug in a=1, b=4, and c=-6




{{{u = (-4 +- sqrt( 16-4*1*-6 ))/(2*1)}}} Square 4 to get 16  




{{{u = (-4 +- sqrt( 16+24 ))/(2*1)}}} Multiply {{{-4*-6*1}}} to get {{{24}}}




{{{u = (-4 +- sqrt( 40 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{u = (-4 +- 2*sqrt(10))/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{u = (-4 +- 2*sqrt(10))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{u = (-4 + 2*sqrt(10))/2}}} or {{{u = (-4 - 2*sqrt(10))/2}}}



Now break up the fraction



{{{u=-4/2+2*sqrt(10)/2}}} or {{{u=-4/2-2*sqrt(10)/2}}}



Simplify



{{{u=-2+sqrt(10)}}} or {{{u=-2-sqrt(10)}}}



Remember, we let {{{u=4^x}}}. So 


{{{4^x=-2+sqrt(10)}}} or {{{4^x=-2-sqrt(10)}}}



Let's solve the first equation {{{4^x=-2+sqrt(10)}}} 


{{{4^x=-2+sqrt(10)}}} Start with the first equation


{{{log(10,(4^x))=log(10,(-2+sqrt(10)))}}} Take the log of both sides


{{{x*log(10,(4))=log(10,(-2+sqrt(10)))}}} Rewrite the left side using the identity  {{{log(b,(x^y))=y*log(b,(x))}}}


{{{x=log(10,(-2+sqrt(10)))/log(10,(4))}}} Divide both sides by {{{log(10,(4))}}} to isolate x


{{{x=log(4,(-2+sqrt(10)))}}} Combine the logs by use of the change of base formula



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Now let's solve the second equation {{{4^x=-2-sqrt(10)}}} 


{{{4^x=-2+sqrt(10)}}} Start with the first equation


{{{log(10,(4^x))=log(10,(-2-sqrt(10)))}}} Take the log of both sides



Notice how {{{-2-sqrt(10)=-5.16228}}}. Since we cannot take the log of a negative number, this means we must ignore it.







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Answer:

So the solution is 


{{{x=log(4,(-2+sqrt(10)))}}}


which approximates to 


{{{x=0.108477}}}