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Basic of Logarithms
hard
Let $\log _a b=4, \log _c d=2$, where $a, b, c, d$ are natural numbers. Given that $b-d=7$, the value of $c-a$ is
A
$1$
B
$-1$
C
$2$
D
$-2$
(KVPY-2009)
Solution
(a)
We have, $\log _a b=4$, $\log _c d=2, a, b, c, d \in N$
$\Rightarrow b=a^4, d=c^2$
$\Rightarrow b-d=7=a^4-c^2$
$\Rightarrow 7=\left(a^2+c\right)\left(a^2-c\right)$
$\Rightarrow 7 \times 1=\left(a^2+c\right)\left(a^2-c\right)$
$\therefore a^2+c=7 \text { and } a^2-c=1$
$\text { On solving, we get } a=2 \text { and } c=3$
$\therefore c-a=3-2=1$
Standard 11
Mathematics
