Question

The non zero root of the equation 3^2x+9=10.3^x is :-

a) a +ve fraction

b) a -ve fraction

c) a +ve integer

d) a -ve integer

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{equation}\mathrm{is}, \\ {3}^{2x}+9=10.3^x \\ \mathrm{put}{3}^x=y,\mathrm{then}\mathrm{equation}\mathrm{becomes}, \\ {y}^2+9-10y=0 \\ \Rightarrow y^2-10y+9=0 \\ \Rightarrow y^2-9y-y+9=0 \\ \Rightarrow y(y-9)-1(y-9)=0 \\ \Rightarrow (y-1)(y-9)=0 \\ \Rightarrow y=1ory=9 \\ WHENy=1: \\ {3}^x=1 \\ \Rightarrow 3^x=3^0 \\ \Rightarrow x=0 \\ WHENy=9 \\ {3}^x=9 \\ \Rightarrow 3^x=3^2 \\ \Rightarrow x=2 \end{gathered}$$
So, the non zero root of the equation is 2 that is a positive integer.