Question

Let $S$ be the set of all real roots of the equation, \(3^x(3^x-1)+2=\left | 3^x-1 \right |+\left | 3^x-2 \right|\). Then $S$ :

Answer 1

\(3^x(3^x-1)+2=\left | 3^x-1 \right |+\left | 3^x-2 \right |\)

Let \(3^x=t\)

Then \(t(t-1)+2=\left | t-1 \right |+\left | t-2 \right |\)
\(\Rightarrow t^2-t+2=\left | t-1 \right |+\left | t-2 \right |\)

We plot \(t^2-t+2\) and \(\left | t-1 \right |+\left | t-2 \right |\)
As \(t=3^x\) is always positive, therefore only positive values of \(t\) will be the solution.



The graphs intersect for only positive value of \(t\).

So, there is single value of \(x\) given by \(x= \log_3 t\)

Therefore, we have only one solution.