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Moore's law says that the number of transistors in a dense integrated circuit increases by 
41% every year. In 1974, a dense integrated circuit was produced with 5000 transistors.
Which expression gives the number of transistors in a dense integrated circuit in 1979?
Choose 1 answer:
(A) 
5000+0.41^(5)
(B) 
5000*0.41^(5)
(C) 
5000(1+0.41)^(5)
(D) 
5000+(1+0.41)^(5)

Moore's law says that the number of transistors in a dense integrated circuit increases by 41% 41 \% every year. In 19741974, a dense integrated circuit was produced with 50005000 transistors.\newlineWhich expression gives the number of transistors in a dense integrated circuit in 19791979?\newlineChoose 11 answer:\newline(A) 5000+0.415 5000+0.41^{5} \newline(B) 50000.415 5000 \cdot 0.41^{5} \newline(C) 5000(1+0.41)5 5000(1+0.41)^{5} \newline(D) 5000+(1+0.41)5 5000+(1+0.41)^{5}

Full solution

Q. Moore's law says that the number of transistors in a dense integrated circuit increases by 41% 41 \% every year. In 19741974, a dense integrated circuit was produced with 50005000 transistors.\newlineWhich expression gives the number of transistors in a dense integrated circuit in 19791979?\newlineChoose 11 answer:\newline(A) 5000+0.415 5000+0.41^{5} \newline(B) 50000.415 5000 \cdot 0.41^{5} \newline(C) 5000(1+0.41)5 5000(1+0.41)^{5} \newline(D) 5000+(1+0.41)5 5000+(1+0.41)^{5}
  1. Understanding Moore's Law: Understand Moore's Law.\newlineMoore's Law suggests that the number of transistors on a dense integrated circuit doubles approximately every two years. However, in this problem, we are given a yearly increase of 41%41\%. This means that each year, the number of transistors is 141%141\% of the previous year's number.
  2. Converting the percentage increase: Convert the percentage increase to a decimal multiplier.\newlineAn increase of 41%41\% per year means that the number of transistors is multiplied by 1.411.41 (which is 1+0.411 + 0.41) each year.
  3. Determining the expression for 19791979: Determine the expression for the number of transistors in 19791979.\newlineSince the increase is compounded annually, we need to use an exponential growth formula. The initial number of transistors is 50005000, and the growth happens over 55 years (from 19741974 to 19791979). The expression for the number of transistors in 19791979 is therefore 5000×(1.41)55000 \times (1.41)^5.
  4. Matching the expression with options: Match the expression with the given options.\newlineThe correct expression from Step 33 is 5000×(1.41)55000 \times (1.41)^5, which is equivalent to 5000×(1+0.41)55000 \times (1 + 0.41)^5. This matches option (C) 5000(1+0.41)55000(1+0.41)^{5}.

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