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Logarithmic Differentiation

If a, b and...

Question

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

- 1

and

\frac{{log}_{e} c}{{log}_{e} a}

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1

and

\frac{{log}_{e} c}{{log}_{e} a}

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1

and

{log}_{a} c

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- 1

and

{log}_{c} a

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Solution

The correct options are
B $1$ and $\frac{{log}_{e} c}{{log}_{e} a}$
C $1$ and ${log}_{a} c$
Since, $a, b$ and $c$ are in GP.
$∴ b^{2} = a c$

Given equation is $({log}_{e} a) x^{2} - 2 ({log}_{e} b) x + ({log}_{e} c) = 0$

Put $x = 1$ , we get

${log}_{e} a - 2 {log}_{e} b + {log}_{e} c = 0$

$\Rightarrow 2 {log}_{e} b = {log}_{e} a + {log}_{e} c$

$\Rightarrow {log}_{e} b^{2} = {log}_{e} a c$

$\Rightarrow b^{2} = a c$ , which is true.

Hence, one of the root of the given equation is $1$ .

Let another root be $α$ .

$∴$ Sum of roots, $1 + α = \frac{2 {log}_{e} b}{{log}_{e} a} = \frac{{log}_{e} b^{2}}{{log}_{e} a}$

$\Rightarrow α = \frac{{log}_{e} a c}{{log}_{e} a} - 1$

$= \frac{({log}_{e} a + {log}_{e} c)}{{log}_{e} a} - 1$

$= \frac{{log}_{e} c}{{log}_{e} a} = {log}_{a} c$

Hence, roots are $1$ and ${log}_{a} c$

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If a,b and c are positive numbers in a GP, then the roots of the quadratic equation (logea)x2−(2logeb)x+(logec)=0 are

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