Question

The range of values of $x$ that satisfy the inequation $(4^{x+1}-{16}^x<2{\log }_{4}8)$ is

• A. $x\in (0,{\log }_{4}3)$
• B. $x\in (0,{\log }_{3}4)$
• C. $x\in (0,{\log }_{4}3]$
• D. None of these

Answer 1

Answer: (A)

$x\in (0,{\log }_{4}3)$

Given,

$(4^{x+1}-{16}^x<2{\log }_{4}8)$

Now,

$lo{g}_{4}8=lo{g}_4{2}^3=3lo{g}_{4}2=\frac{3}{lo{g}_{2}4}=\frac{3}{2lo{g}_{2}2}=\frac{3}{2}$.............(1)

Let,

$4^x=t$ then given inequation reduce in the form

$4t-t^2>2.\frac{3}{2}$..........from(1)

$\Rightarrow t^2-4t+3<0$

$\Rightarrow (t-1)(t-3)<0$

$\Rightarrow 1<t<3$

$\Rightarrow 1<4^x<3$

$\Rightarrow 0<x<lo{g}_{4}3$

Hence, the answer is option $A$