```Question 143993

{{{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":

{{{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)}}}

{{{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)}}}

{{{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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