Let $l o g_{a} b = 4, l o g_{c} d = 2$ where $a, b, c, d$ are natural numbers. Given that $b - d = 7$ , then value of $c - a$ is

1

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Solution

The correct option is C $1$
$l o g_{a} b = 4$ & $l o g_{c} d = 2$
$b = a^{4} d = c^{2}$
$a^{4} - c^{2} = 7$
$(a^{2} - c) (a^{2} + c) = 7$
$a^{2} - c = 7$ & $a^{2} + c = 1$ not possible
or $a^{2} - c = 1$ & $a^{2} + c = 7$
$2 a^{2} = 8$
$a = \pm 2$
$a = 2 c = 3$
$∴ c - a = 1$

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