if $2^{a} > 4^{c}$ and $3^{b}$ > $9^{a}$ and a, b, c all positive, then :

c < a < b

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b < c < a

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c < b < a

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a < b < c

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Solution

The correct option is A c < a < b
$2^{a} > 4^{c}$
$\Rightarrow 2^{a} > (2^{2})^{c}$
$\Rightarrow 2^{a} > 2^{2 c}$
base are same so
$a > 2 c$
$3^{b} > 9^{a}$
$\Rightarrow 3^{b} > (3^{2})^{a}$
$\Rightarrow b > 2 a$
Hence $b > a > c$

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