If a, b and c are positive numbers in a GP,th | vedclass

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(C) Since $a, b$ and $c$ are in $GP$,we have $b^{2} = ac$. Taking the natural logarithm on both sides,we get $2 \log_{e} b = \log_{e} a + \log_{e} c$.

Vedclass · Accepted Answer

$1$ and $\log_{a} c$

Vedclass · Answer

(C) Since $a, b$ and $c$ are in $GP$,we have $b^{2} = ac$. Taking the natural logarithm on both sides,we get $2 \log_{e} b = \log_{e} a + \log_{e} c$. Given the quadratic equation $(\log_{e} a) x^{2} - (2 \log_{e} b) x + \log_{e} c = 0$. Substituting $x = 1$ into the equation,we get $(\log_{e} a) - (2 \log_{e} b) + \log_{e} c = 0$,which simplifies to $\log_{e} a + \log_{e} c = 2 \log_{e} b$,which is true. Thus,$x = 1$ is one root of the equation. Let the other root be $\alpha$. The product of the roots is given by $\frac{\log_{e} c}{\log_{e} a}$. Therefore,$1 	imes \alpha = \frac{\log_{e} c}{\log_{e} a} = \log_{a} c$. Hence,the roots are $1$ and $\log_{a} c$.

If $a, b$ and $c$ are positive numbers in a $GP$,then the roots of the quadratic equation $(\log_{e} a) x^{2} - (2 \log_{e} b) x + (\log_{e} c) = 0$ are

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