If ${log}_{a} x > y$ and $0 < a < 1$ . Then

$x > y$

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$x > a^{y}$

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$x < a^{y}$

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$x = a^{y}$

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Solution

The correct option is C
$x < a^{y}$

Given: ${log}_{a} x > y$ and $0 < a < 1$

We know the graph of ${log}_{a}$ x if $0 < a < 1$ is decreasing function.

If value of x increases function values decreases

If ${log}_{a} n > y$

${log}_{a} x > {log}_{a} a^{y}$

Above given inequality is true only when

$a^{y} > x$

Suggest Corrections

Similar questions

If ${log}_{a} x > y$ and $a > 1$ . Then

y = a^{\frac{1}{1 - {log}_{a} x}}

z = a^{\frac{1}{1 - {log}_{a} y}}

x =

If ${log}_{a} x > y$ and $0 < a < 1$ . Then

- 1 :

0 < x < y \Rightarrow {log}_{a} x > {log}_{a} y

a > 1

- 2 :

a > 1, {log}_{a} x

y = l o g_{a} x

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