Question

find the quadratic polynomial whose zeroes are log5 to base25 log3 to base 27​

Answer 1

Step-by-step explanation:

Let the zeros of the quadratic polynomial be $\alpha \: \beta$

$\therefore \: \alpha = log_{25} 5 = \frac{log \: 5}{log \: 25} \\ = \frac{log \: 5}{log \: {5}^{2} } \\= \frac{log \: 5}{2 \: log \: 5} = 2 \\ \implies \: \alpha = 2 \\ \beta = log_{27} 3 = \frac{log \: 3}{log \: 27} \\ = \frac{log \: 3}{log \: {3}^{3} } \\= \frac{log \: 3}{3 \: log \: 3} = 3\\ \implies \: \beta = 2 \\ now \: required \: polynomial \: is : \\ {x}^{2} - ( \alpha + \beta )x + \alpha \beta \\ = {x}^{2} - (2 + 3)x + 2 \times 3 \\ = {x}^{2} - 5x + 6$