AMC10 2003 B

AMC10 2003 B · Q9

AMC10 2003 B · Q9. It mainly tests Manipulating equations, Primes & prime factorization.

Find the value of $x$ that satisfies the equation $25^{-2} = 548/x \div 526/x \cdot 2517/x$.

求满足方程 $25^{-2} = 548/x \div 526/x \cdot 2517/x$ 的 $x$ 的值。

(A) 2 2

(B) 3 3

(D) 6 6

(E) 9 9

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) Write all the terms with the common base 5. Then $5^{-4}=25^{-2}=\dfrac{5^{48/x}}{5^{26/x}\cdot25^{17/x}}=\dfrac{5^{48/x}}{5^{26/x}\cdot5^{34/x}}=5^{(48-26-34)/x}=5^{-12/x}.$ It follows that $-\dfrac{12}{x}=-4$, so $x=3$. OR First write $25$ as $5^2$. Raising both sides to the $x$ power gives $5^{-4x}=\dfrac{5^{48}}{5^{26}5^{34}}=5^{48-26-34}=5^{-12}.$

（B）把所有项都写成以 5 为底的形式。则 $5^{-4}=25^{-2}=\dfrac{5^{48/x}}{5^{26/x}\cdot25^{17/x}}=\dfrac{5^{48/x}}{5^{26/x}\cdot5^{34/x}}=5^{(48-26-34)/x}=5^{-12/x}.$ 因此 $-\dfrac{12}{x}=-4$，所以 $x=3$。或者先把 $25$ 写成 $5^2$。两边同取 $x$ 次幂，得 $5^{-4x}=\dfrac{5^{48}}{5^{26}5^{34}}=5^{48-26-34}=5^{-12}.$

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