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Zorluk: ZorExponents, Radicals, and Algebraic Expressions

If aa and bb are positive integers such that 2a+35b2a5b+1=6,0002^{a+3} \cdot 5^b - 2^a \cdot 5^{b+1} = 6,000, what is the value of aba \cdot b?

  1. A
    7
  2. B
    10
  3. 12Cevap
  4. D
    15
  5. E
    16

Cevap

12
Factoring out 2a5b2^a \cdot 5^b yields 2a5b(2351)=32a5b=6,0002^a \cdot 5^b (2^3 - 5^1) = 3 \cdot 2^a \cdot 5^b = 6,000. Dividing by 3 yields 2a5b=2,0002^a \cdot 5^b = 2,000. Expressing 2,0002,000 in prime factored form gives 24532^4 \cdot 5^3. Matching exponents for the prime bases gives a=4a = 4 and b=3b = 3. The product aba \cdot b is 4×3=124 \times 3 = 12.

Adım Adım Çözüm

1
Rewrite the terms in the expression using exponent rules to isolate common bases.
2a+35b2a5b+1=(2a23)5b2a(5b51)2^{a+3} \cdot 5^b - 2^a \cdot 5^{b+1} = (2^a \cdot 2^3) \cdot 5^b - 2^a \cdot (5^b \cdot 5^1)
Applying xm+n=xmxnx^{m+n} = x^m \cdot x^n allows us to extract common powers of 2a2^a and 5b5^b.
2
Factor out the common term 2a5b2^a \cdot 5^b from the left side of the equation.
2a5b(2351)=2a5b(85)=32a5b2^a \cdot 5^b (2^3 - 5^1) = 2^a \cdot 5^b (8 - 5) = 3 \cdot 2^a \cdot 5^b
Simplifying the constant factor in parentheses simplifies the equation.
3
Divide both sides of the equation by 3 and perform prime factorization on the resulting integer.
32a5b=6,000    2a5b=2,000=24533 \cdot 2^a \cdot 5^b = 6,000 \implies 2^a \cdot 5^b = 2,000 = 2^4 \cdot 5^3
Prime factorization of 2,0002,000 determines the unique integer exponents for bases 2 and 5.
4
Equate the corresponding exponents and calculate the requested product aba \cdot b.
a=4a = 4 and b=3    ab=4×3=12b = 3 \implies a \cdot b = 4 \times 3 = 12
Since 2 and 5 are prime numbers, the prime factorization representation is unique.

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Factoring Exponents and Prime Factorization
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