Question

Q8.
Ans: 0
My answer : In LHS , whatever value of x, 1- root x-1 will either come 0 or a negative value.
Since this is a limitation of log that nber can't be 0 and should be greater than 0. Therefore this equation not possible.
Is this correct?
Q.8. If ${\log }_a\left(1-\sqrt{1+x}\right)={\log }_{a^2}\left(3-\sqrt{1+x}\right)$, then number of solutions of the equation is-
(A) 0
(B) 1
(C) 2
(D) infinitely many

Answer 1

Dear student
$$\begin{gathered} {\log }_{\mathrm{a}}\left(1-\sqrt{1+\mathrm{x}}\right)={\log }_{{\mathrm{a}}^2}\left(3-\sqrt{1+\mathrm{x}}\right) \\ \Rightarrow \frac{\log \left(1-\sqrt{1+\mathrm{x}}\right)}{\mathrm{loga}}=\frac{\log \left(3-\sqrt{1+\mathrm{x}}\right)}{{\mathrm{loga}}^2}\left[\mathrm{As}{\log }_{\mathrm{a}}\mathrm{b}=\frac{\mathrm{logb}}{\mathrm{loga}}\right] \\ \Rightarrow \frac{\log \left(1-\sqrt{1+\mathrm{x}}\right)}{\mathrm{loga}}=\frac{\log \left(3-\sqrt{1+\mathrm{x}}\right)}{2\mathrm{loga}}\left[\mathrm{As}\log \left({\mathrm{x}}^{\mathrm{a}}\right)=\mathrm{alog}\left(\mathrm{x}\right)\right] \\ \Rightarrow 2\log \left(1-\sqrt{1+\mathrm{x}}\right)=\log \left(3-\sqrt{1+\mathrm{x}}\right) \\ \Rightarrow \log {\left(1-\sqrt{1+\mathrm{x}}\right)}^2=\log \left(3-\sqrt{1+\mathrm{x}}\right) \\ \Rightarrow {\left(1-\sqrt{1+\mathrm{x}}\right)}^2=3-\sqrt{1+\mathrm{x}}\left[\mathrm{As}\log \left(\mathrm{f}\right(\mathrm{x}\left)\right)=\log \left(\mathrm{g}\right(\mathrm{x}\left)\right)\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right)\right] \\ \Rightarrow 1+1+\mathrm{x}-2\sqrt{1+\mathrm{x}}=3-\sqrt{1+\mathrm{x}} \\ \Rightarrow -1+\mathrm{x}=\sqrt{1+\mathrm{x}} \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow 1+{\mathrm{x}}^2-2\mathrm{x}=1+\mathrm{x} \\ \Rightarrow {\mathrm{x}}^2-3\mathrm{x}=0 \\ \Rightarrow \mathrm{x}(\mathrm{x}-3)=0 \\ \Rightarrow \mathrm{x}=0\mathrm{or}\mathrm{x}-3=0 \\ \Rightarrow \mathrm{x}=0\mathrm{or}\mathrm{x}=3 \\ \mathrm{So},\mathrm{there}\mathrm{are}2\mathrm{solutions}. \end{gathered}$$
Regards