NIMCET 2018 Question Paper with solutions

As, $\mathit{l}\mathit{o}{\mathit{g}}_3$ 2 , $\mathit{l}\mathit{o}{\mathit{g}}_3\left(2^{\mathit{x}}-5\right)$ , $\mathit{l}\mathit{o}{\mathit{g}}_3\left(2^{\mathit{x}}-\frac{7}{2}\right)$ are in arithmetic progression,
∴ $\mathit{l}\mathit{o}{\mathit{g}}_3\left(2^{\mathit{x}}-5\right)$ - $\mathit{l}\mathit{o}{\mathit{g}}_{3}2$
= $\mathit{l}\mathit{o}{\mathit{g}}_3\left(2^{\mathit{x}}-\frac{7}{2}\right)-\mathit{l}\mathit{o}{\mathit{g}}_3\left(2^{\mathit{x}}-5\right)$
= $\mathit{l}\mathit{o}{\mathit{g}}_3\left(\frac{2^{\mathit{x}}-5}{2}\right)$ = $\mathit{l}\mathit{o}{\mathit{g}}_3\left(\frac{2^{\mathit{x}}-\frac{7}{2}}{2^{\mathit{x}}-5}\right)$ $\left[\mathrm{Since}\log \left(\frac{\mathit{a}}{\mathit{b}}\right)=\mathit{l}\mathit{o}\mathit{g}\mathit{a}-\mathit{l}\mathit{o}\mathit{g}\mathit{b}\right]$
⇒ $\left(2^{\mathit{x}}-5\right)\left(2^{\mathit{x}}-5\right)$ = $\left(2^{\mathit{x}}-\frac{7}{2}\right)$ × 2
⇒ ${\left(2^{\mathit{x}}-5\right)}^2$ = $2^{\mathit{x}+1}$ - 7
⇒ $2^{2\mathit{x}}$ + 25 - 2 × $2^{\mathit{x}}$ × 5 = $2^{\mathit{x}+1}$ - 7
⇒ $2^{\mathit{x}+1}+2^{\mathit{x}+1}$ × 5 - $2^{2\mathit{x}}$ = 32
⇒ $2^{\mathit{x}-1}$ (6) - $2^{2\mathit{x}}$ = 32
⇒ $2^{\mathit{x}}\left(12-2^{\mathit{x}}\right)$ = 32
The above equation is satisfied for x = 3