Question

if log 81 base 2 and log 8 base to add zeros of polynomial then find the polynomial​

Answer 1

Given :

Roots or zeroes of the required polynomial = $\log_281$ and $\log_2 8$

To Find :

What is the polynomial having the given roots ?

Solution :

We know that the equation of a polynomial is given as :

$x^2 -$ ( sum of the two  roots of required equation ) .$x$ + ( product of the two roots of required equation  )  = 0

Or , $x^2 -$ ( $\log_281$ + $\log_2 8$) .$x$ +(

Or ,  $x^2 -$ ( $\log_2 (81\times 8)$) .$x$ + ($(log_281 )^3$) = 0    [∴loga + logb = log(ab) & $\log_2 8$ =3]

Or ,  $x^2 -$ $log_2(648).x$ + $3 \log_2 81$ = 0

Hence , the required polynomial is $x^2 -$ $log_2(648).x$ + $3 \log_2 81$ = 0 .