Question 1143119
{{{r (x) = 4^(2x+1) - 8}}}

{{{r (x) = (2^2)^(2x+1) - 8}}}

{{{r (x) = 2^(4x+2) - 8}}}


{{{(d/dx)(2^(4x+2) - 8)=  (d/dx)(2^(4x+2))-(d/dx)(8))}}}.....apply exponent rule

{{{(d/dx)(2^(4x+2) - 8)=log(2)*2^(4x + 2)*(d/dx)(4x + 2) -0}}}

{{{(d/dx)(2^(4x+2) - 8)=ln(2)(4*(d/dx)(x)+(d/dx)(2))*2^(4x+2)}}}

{{{(d/dx)(2^(4x+2) - 8)=ln(2)(4*1+0)*2^(4x+2)}}}

{{{(d/dx)(2^(4x+2) - 8)=ln(2)*4*2^(4x+2)}}}

{{{(d/dx)(2^(4x+2) - 8)=ln(2)*2^2*2^(4x+2)}}}

{{{(d/dx)(2^(4x+2) - 8)=ln(2)*2^(4x+2+2)}}}

{{{(d/dx)(2^(4x+2) - 8)=ln(2)*2^(4x+4)}}}



solve the equation {{{r (x) = 0}}}

{{{4^(2x+1) - 8=0}}}

{{{4^(2x+1) =8}}}

{{{2^(4x+2) =2^3}}}...if bases same, exponents are same too

{{{4x+2 =3}}}

{{{4x=3-2}}}

{{{4x=1}}}

{{{x=1/4}}}