Question

If $ \log_2 (9 ^{x-1} + 7) - \log_2 (3^{x-1 } + 1) = 2 $ then $x$ values are

• A. 0,2
• B. 1,3
• C. 1,4
• D. 1,2

Answer 1

Answer: (D)

$log_{2} \left(9^{x-1} +7\right)-log_{2}\left(3^{x-1} +1\right) = 2$
$\Rightarrow \,log_{2}\left(\frac{9^{x-1} +7}{3^{x-1} +1}\right) = 2 \,log_{2}\, 2$
$\Rightarrow log_{2}\left(\frac{9^{x-1} +7}{3^{x-1} +1}\right) = log_{2} 2^{2}$
$\Rightarrow \left(\frac{9^{x-1} +7}{3^{x-1} +1}\right) = 4$
$\Rightarrow \frac{\left(3^{2}\right)^{x-1} +7}{\left(3^{x-1} +1\right)} = 4$
$\Rightarrow \frac{\left(3^{2}\right)^{x-1} +7}{3^{x-1} +1} = 4$
Let $3^{x-1} = y$
$\therefore \frac{y^{2}+7}{y+1} = 4$
$\Rightarrow y^{2}+7=4y+4$
$\Rightarrow y^{2}-4y+3 = 0$
$\Rightarrow y^{2} -3y-y+3 = 0$
$\Rightarrow y\left(y-3\right)-1\left(y-3\right) = 0$
$\Rightarrow \left(y-3\right)\left(y-1\right) = 0$
$\Rightarrow y = 3,1$
If y = 3, then
$3^{x-1} = 3$
$\Rightarrow x-1 = 1$
$ \Rightarrow x = 2$
If y = 1, then
$3^{x-1} = 3^{0}$
$\Rightarrow x-1 = 0$
$ \Rightarrow x = 1$
$\therefore x =1, 2$