If $\alpha =\log _{ 3 }{ 4 }$ is one of the roots of equation ${ x }^{ 3 }+a{ x }^{ 2 }+bx+1=0$, $a,b\in R$, then equation which always have a root $\beta =\log _{ 2 }{ 48 }$, is
- ${ x }^{ 3 }+2\left( b-6 \right) { x }^{ 2 }+4\left( a-4b+12 \right) x+8\left( 4b-2a-7 \right) =0$
- ${ x }^{ 3 }+b{ x }^{ 2 }+ax+1=0$
- $\left( 65+16a+4b \right) { x }^{ 2 }+2\left( 48+8a+b \right) { x }^{ 2 }+4\left( 12+a \right) x+8=0$
- $\left( 65+16a+4b \right) { x }^{ 2 }+\left( 48+8a+b \right) { x }^{ 2 }+\left( 12+a \right) x+8=0 $
Let t = log2(3). Then alpha = 2/t and beta = 4 + t, so alpha(beta - 4) = 2. Substituting alpha = 2/(beta - 4) into the given cubic produces the equation in option A.
Given that alpha = log_3(4), the term beta = log_2(48) can be rewritten using base 3 as beta = log_3(48) / log_3(2). We can simplify the logarithms by using the fact that log_3(4) = 2*log_3(2), which means log_3(2) equals alpha/2. We also know that log_3(48) equals log_3(16) + log_3(3), which simplifies to 4*log_3(2) + 1, or 2*alpha + 1. Substituting these into the expression for beta gives beta = (2*alpha + 1) / (alpha/2) = 4 + 2/alpha. Since alpha is a root of x^3 + ax^2 + bx + 1 = 0, dividing the entire equation by x yields alpha^2 + a*alpha + b + 1/alpha = 0, meaning 1/alpha equals -(alpha^2 + a*alpha + b). Substituting this back results in beta = 4 - 2*(alpha^2 + a*alpha + b), and substituting this entire expression for x into the provided correct equation algebraically confirms it equals zero.