JEE MainMathematicsToughLogarithm and its applications

The number of positive integers not satisfying the inequality
$\log_{2}\left( 4^{x} - 2 \cdot 2^{x} + 17 \right) > 5$

Show answer and solution

\log_{2}{(4^{x} - {2.2}^{x} + 17)} > 5

\Rightarrow 4^{x} - {2.2}^{x} + 17 > 2^{5}

Put

2^{x} = t

\Rightarrow t^{2} - 2t - 15 > 0

\Rightarrow (t + 3)(t - 5) > 0

t < - 3

t > 5

t < - 3

rejected at

2^{x}

can't be negative.
If if does not satisfy

2^{x} > 5

then it must satisfy

2^{x} \leq 5

\Rightarrow x = 1;\ \ \ 2^{1} \leq 5

x = 2;\ \ \ \ \ \ \ \ \ 2^{2} = 4 \leq 5

x = 3;\ \ \ \ \ \ \ \ \ 2^{3} = 8 \nleq 5

\Rightarrow x = 1

and 2

Source: Grass-G / M-05 Mathematics Paper 10 October.docx

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