Question

What is the sum of the roots of the equation $4^x-3\left(2^{x+3}\right)+128=0$?

Answer 1

Given equation

$4^x-3\left(2^{x+3}\right)+128=0$

$\Rightarrow (2^2{)}^x-3(2^x{.2}^3)+128=0$

$\Rightarrow (2^x{)}^x-3(2^x.8)+128=0$

Taking $2^x=y$ the equation becomes

$\Rightarrow y^2-24y+128=0$

$\Rightarrow y^2-16y-8y+128=0$

$\Rightarrow y(y-16)-8(y-16)=0$

So $y=8$ and $y=16$

when $y=8\Rightarrow 2^x=2^x\Rightarrow x=3$

when $y=16\Rightarrow 2^x=2^4\Rightarrow x=4$

Hence roots are $3$ and $4$

So teh sum of the roots is $=3+4=7$