University:
Cornell University
Course:
MATH 1101 | Calculus Preparation
Academic year:

2020
Views:

198

Pages:

2
Author:

Clint W.

(9m^4 - 12m^2) / (3m^2) = 9m^(4-2) - 12m^(2-2) = 3m^2 - 4 2. (25m^5 - 30m^2) / (5m^2) = 5m^(5-2) - 6m^(2-2) = 5m^3 - 6 3. (6 + x^-2)^-2 = (6 + 1/x^2)^-2 = (x^2/6x^2 + 1)^-2 = (x^2/(6x^2 + 1))^2 = x^4 / (36x^4 + 12x^2 + 1) 4. (5^-2 + 5^-3)^-1 = (1/25 + 1/125)^-1 = (5/125 + 1/125)^-1 = (6/125)^-1 = 125/6 5. (216^-3 * 36) / (6^-5) = (6^3)^-3 * 6^2 / 6^-5 = 6^-9 * 6^2 / 6^-5 = 6^-7 / 6^-5 = 1/6^2 = 1/36 6. (-3n)^7 / (-3n)^-3 = (-3n)^(7+3) = (-3)^10 * n^10 = 3^10 * n^10 7. -x^2y^-5 * 3x^3y^7 = -3x^(2+3) * y^(-5+7) = -3x^5 * y^2 8. (3x^-3)^2 * (-3x^4)^-4 = (3^2 * x^-6) * ((-3)^-4 * x^-16) = (9 * x^-6) * (1/(-3)^4 * x^-16) = 9 / (3^4 * x^22) = 1 / (9 * x^22) 9. (27a^4)^(2/3) / (27a^4)^(-1/3) = (27a^4)^(2/3 + 1/3) = (27a^4)^1 = 27a^4 10. 6ab^(1/3) / (3ab^(-1/2)) = 2a^(1-1) * b^(1/3 + 1/2) = 2a^0 * b^(5/6) = 2 * b^(5/6) 11. 4y^(3/2) / (2y^(1/2) - 3y^(-1/2)) = 4y^(3/2 - 1/2) / (2 - 3/y) = 4y / (2 - 3/y) = 8y / (8 - 12/y) 12. √(64a^-9b^5c^6) = 4a^(-9/2) * b^(5/2) * c^(6/2) = 4b^(5/2) * c^3 / a^3 Solve the following exponential equations for x. Remember to include and check all possible answers! 13. 9^(x+1) = √27 (3^2)^(x+1) = √(3^3) 3^(2x+2) = 3^(3/2) 2x+2 = 3/2 2x = -1/2 x = -1/4 14. (4+x)^-2 = 256 4+x = (256)^(-1/2) 4+x = ±1/16 x = 1/16 - 4, -1/16 - 4 x = -63/16, -65/16 15. (1/2)^(x-3) = 32^(x+1) (2^-1)^(x-3) = (2^5)^(x+1) 2^(-x+3) = 2^(5x+5) -x+3 = 5x+5 -6 = 6x x = -1 16. 3^(x-3) = 9 3^(x-3) = 3^2 x-3 = 2 x = 5 17. 5^(u-6) + 5^(u+5) = 0 Let u = 5^x u(u-6) + 5(u-6) = 0 (u-6)(u+5) = 0 u = 6 or u = -5 5^x = 6, 5^x = -5 (no solution) x = log₅(6) 18. 3^(x²+3x) = 81 3^(x²+3x) = 3^4 x²+3x = 4 x²+3x-4 = 0 (x+4)(x-1) = 0 x = -4, x = 1 19. The price of firewood four years ago was $140 per cord. Today, a cord of wood costs $182. To the nearest tenth of a percent, what has been the annual rate of increase in the cost? 182 = 140(1+r)^4 1.3 = (1+r)^4 (1.3)^(1/4) = 1+r r = (1.3)^(1/4) - 1 ≈ 6.8% 20. A man invests $5,000 in an account that pays 8.5% interest per year, compounded quarterly. Find the amount after 3 years. 5000(1+0.085/4)^(4*3) ≈ $6,435.09 21. A sum of $1,000 was invested for 4 years, and the interest was compounded semiannually. If this sum amounted to $1,435.77 in the given time, what was the interest rate? 1435.77 = 1000(1+r/2)^(2*4) 1.43577 = (1+r/2)^8 (1.43577)^(1/8) - 1 = r/2 r ≈ 1.25% 22. Find the balance when $16,000 is invested at an annual rate of 8% for 3 years if the interest is compounded monthly? 16000(1+0.08/12)^(12*3) ≈ $20,323.79 23. Find the balance when $19,000 is invested at an annual rate of 6% for 5 years if the interest is compounded daily? 19000(1+0.06/365)^(365*5) ≈ $25,646.68 24. A baseball card increased in value from $50 to $500 in 15 years. Find its average annual rate of appreciation. 500 = 50(1+r)^15 10 = (1+r)^15 10^(1/15) = 1+r r ≈ 16.4% 25. A new car that cost $12,000 decreased in value to $4,000 in 5 years. Find the average annual rate of depreciation. 4000 = 12000(1-r)^5 1/3 = (1-r)^5 (1/3)^(1/5) = 1-r r = 1 - (1/3)^(1/5) ≈ 13.7%

Math Review of Exponents Quiz Review 1 Solution

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