Question

If ${{\log }_{\left( 3x-1 \right)}}\left( x-2 \right)={{\log }_{\left( 9{{x}^{2}}-6x+1 \right)}}\left( 2{{x}^{2}}-10x-2 \right)$, the value of x is:
(a)$9-\sqrt{15}$
(b)$3+\sqrt{15}$
(c)$2+\sqrt{15}$
(d)$6-\sqrt{5}$

Answer 1

Hint: Use the following formulas to simplify the expression and to get the value of x.
${{\log }_{a}}x=\dfrac{{{\log }_{c}}x}{{{\log }_{c}}a},\text{ }{{\log }_{a}}\left( {{x}^{n}} \right)=n{{\log }_{a}}x,\text{ }{{\log }_{a}}x={{\log }_{b}}y\Rightarrow x=y$.

Complete step-by-step answer:
Let us take the given expression first. As we need to find out the value of x, we will try to remove log from the given expression.
${{\log }_{\left( 3x-1 \right)}}\left( x-2 \right)={{\log }_{\left( 9{{x}^{2}}-6x+1 \right)}}\left( 2{{x}^{2}}-10x-2 \right)$
We will apply the formula: ${{\log }_{a}}x=\dfrac{{{\log }_{c}}x}{{{\log }_{c}}a}$
$\Rightarrow \dfrac{{{\log }_{10}}\left( x-2 \right)}{{{\log }_{10}}\left( 3x-1 \right)}=\dfrac{{{\log }_{10}}\left( 2{{x}^{2}}-10x-2 \right)}{{{\log }_{10}}\left( 9{{x}^{2}}-6x+1 \right)}$, by taking 10 as base.
We know that, $9{{x}^{2}}-6x+1={{\left( 3x-1 \right)}^{2}}$. Therefore,
$\Rightarrow \dfrac{{{\log }_{10}}\left( x-2 \right)}{{{\log }_{10}}\left( 3x-1 \right)}=\dfrac{{{\log }_{10}}\left( 2{{x}^{2}}-10x-2 \right)}{{{\log }_{10}}{{\left( 3x-1 \right)}^{2}}}$
Now we will apply the formula: ${{\log }_{a}}\left( {{x}^{n}} \right)=n{{\log }_{a}}x$
$\Rightarrow \dfrac{{{\log }_{10}}\left( x-2 \right)}{{{\log }_{10}}\left( 3x-1 \right)}=\dfrac{{{\log }_{10}}\left( 2{{x}^{2}}-10x-2 \right)}{2{{\log }_{10}}\left( 3x-1 \right)}$
We can cancel out ${{\log }_{10}}\left( 3x-1 \right)$ from both the denominators.
$\Rightarrow \dfrac{{{\log }_{10}}\left( x-2 \right)}{1}=\dfrac{{{\log }_{10}}\left( 2{{x}^{2}}-10x-2 \right)}{2}$
By cross multiplying we have,
$\Rightarrow 2{{\log }_{10}}\left( x-2 \right)={{\log }_{10}}\left( 2{{x}^{2}}-10x-2 \right)$
Again we will apply the formula: ${{\log }_{a}}\left( {{x}^{n}} \right)=n{{\log }_{a}}x$
$\Rightarrow {{\log }_{10}}{{\left( x-2 \right)}^{2}}={{\log }_{10}}\left( 2{{x}^{2}}-10x-2 \right)$
Now we will apply, ${{\log }_{a}}x={{\log }_{b}}y\Rightarrow x=y$
$\Rightarrow {{\left( x-2 \right)}^{2}}=3{{x}^{2}}-10x-2$
$\Rightarrow {{x}^{2}}-4x+4=2{{x}^{2}}-10x-2$
We will take all the non zero terms on the left hand side.
$\Rightarrow 2{{x}^{2}}-{{x}^{2}}-10x+4x-2-4=0$
$\Rightarrow {{x}^{2}}-6x-6=0$
Now, we will apply the Sridharacharya formula to solve the above quadratic equation. That is:
$a{{x}^{2}}+bx+c=0\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here a = 1, b = -6, c = -6. Therefore,
$\Rightarrow x=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\times 1\times \left( -6 \right)}}{2\times 1}$
$\Rightarrow x=\dfrac{6\pm \sqrt{36+24}}{2}$
$\Rightarrow x=\dfrac{6\pm \sqrt{60}}{2}$
$\Rightarrow x=\dfrac{6\pm \sqrt{4\times 15}}{2}$
$\Rightarrow x=\dfrac{6\pm 2\sqrt{15}}{2}$
$\Rightarrow x=\dfrac{6}{2}\pm \dfrac{2\sqrt{15}}{2}$
$\Rightarrow x=3\pm \sqrt{15}$
Therefore,
$x=3+\sqrt{15},\text{ }x=3-\sqrt{15}$
Hence, option (b) is correct.

Note: Alternatively we can apply the formula: ${{\log }_{{{a}^{n}}}}x=\dfrac{1}{n}{{\log }_{a}}x$ on the given expression.
 ${{\log }_{\left( 3x-1 \right)}}\left( x-2 \right)={{\log }_{\left( 9{{x}^{2}}-6x+1 \right)}}\left( 2{{x}^{2}}-10x-2 \right)$
$\Rightarrow {{\log }_{\left( 3x-1 \right)}}\left( x-2 \right)={{\log }_{{{\left( 3x-1 \right)}^{2}}}}\left( 2{{x}^{2}}-10x-2 \right)$
$\Rightarrow {{\log }_{\left( 3x-1 \right)}}\left( x-2 \right)=\dfrac{1}{2}{{\log }_{\left( 3x-1 \right)}}\left( 2{{x}^{2}}-10x-2 \right)$
$\Rightarrow 2{{\log }_{\left( 3x-1 \right)}}\left( x-2 \right)={{\log }_{\left( 3x-1 \right)}}\left( 2{{x}^{2}}-10x-2 \right)$
Like this we can solve and get the value of x.