Let log _a b=4 and log _c d=2,where a, b, c, | vedclass

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(A) Given $\log _a b=4$ and $\log _c d=2$,where $a, b, c, d \in \mathbb{N}$.From the definition of logarithms,we have $b=a^4$ and $d=c^2$.Given $b-d=7

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(A) Given $\log _a b=4$ and $\log _c d=2$,where $a, b, c, d \in \mathbb{N}$.From the definition of logarithms,we have $b=a^4$ and $d=c^2$.Given $b-d=7$,we substitute the expressions for $b$ and $d$:$a^4-c^2=7$This can be written as a difference of squares: $(a^2)^2 - c^2 = 7$,which factors as $(a^2-c)(a^2+c)=7$.Since $a, c \in \mathbb{N}$,$a^2+c$ and $a^2-c$ must be factors of $7$. Since $a^2+c > a^2-c$ and $7$ is a prime number,we must have:$a^2+c=7$ and $a^2-c=1$.Adding these two equations: $2a^2 = 8$ $\Rightarrow a^2 = 4$ $\Rightarrow a = 2$ (since $a \in \mathbb{N}$).Substituting $a=2$ into $a^2+c=7$: $4+c=7 \Rightarrow c=3$.Thus,$c-a = 3-2 = 1$.

Let $\log _a b=4$ and $\log _c d=2$,where $a, b, c, d$ are natural numbers. Given that $b-d=7$,the value of $c-a$ is:

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