CAT 2017 Quantitative Ability

$\Rightarrow\;\; \frac{\log(x-3)}{\log3^{2}} = \;\; \frac{\log(x-7)}{\log3}$ $\Rightarrow\;\; \frac{\log(x-3)}{2\log3} = \;\; \frac{\log(x-7)}{\log3}$ $\Rightarrow \log(x-3)=2\log(x-7)$ $\Rightarrow (x-3)=(x-7)^{2}$ $\Rightarrow x-3=x^{2}-14x+49$ $x^{2}-15x+52=0$ Discriminant = $15^{2}-4\times52=225-208=17>0$ Two distinct values of x exists However, since we have the log function involved here, we will have to verify the solutions. Log is defined only for positive terms So. if log(x - 3) and log(x - 7) are defined, x has to be greater than 7. For the equation $x^{2}-15x+52=0$ one of the values $\;\; \frac{1}{2}(15-\sqrt{17})$ This is less than 7. Since x cannot be less than 7 there is only one possible solution