Question

If ${\log }_2(9^{x-1}+7)-{\log }_2(3^{x-1}+1)=2$, then $x$ values are

• A. $0,2$
• B. $0,1$
• C. $1,4$
• D. $1,2$

Answer 1

Answer: (D)

$1,2$

${\log }_2(9^{x-1}+7)-{\log }_2(3^{x-1}+1)=2$

$\Rightarrow {\log }_2\left(\frac{9^{x-1}+7}{3^{x-1}+1}\right)=2$

$\Rightarrow \frac{9^{x-1}+7}{3^{x-1}+1}=4$

$\Rightarrow 9^{x-1}+7={4.3}^{x-1}+4$

$\Rightarrow 3^{2x-2}-{4.3}^{x-1}+3=0$

$\Rightarrow 3^{2x}-{12.3}^x+27=0$

Let $3^x$ be $t$

$\Rightarrow t^2-12t+27=0$

$\Rightarrow t=3,9$

$\Rightarrow 3^x=3^1,3^2$

$\Rightarrow x=1,2$