Question

Solve for $x$:
$3\left(9^x\right)<8\left(3^x\right)+3$

• A. $(-\infty ,1)$
• B. $(0,81)$
• C. None of these.
• D. $(-\infty ,-2)\cup (-2,-1)\cup (-1,0)$

Answer 1

The correct option is A

$(-\infty ,1)$

$\text{Given}:3\left(9^x\right)<8\left(3^x\right)+3$
$\Rightarrow 3(3^x{)}^2<8(3^x)+3$
Put $3^x=t\text{then}t>0)$

$\therefore 3t^2-8t-3<0\Rightarrow (3t+1)(t-3)<0$
$\Rightarrow -\frac{1}{3}<t<3$
But since$t=3^x$ which means, t can never be negative
$\Rightarrow 0<t<3$
$\Rightarrow 0<3^x<3$
$\Rightarrow -\infty <x<1$
$\Rightarrow x\in (-\infty ,1)$
Hence, The correct answer is option B