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    Logarithms and Exponents Applications Word Problems

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    1. Complete table for savings in which interest is compounded continously.

    Initial investment Annual rate Time to double total
    $1000 ? ? $2281.88

    2. Complete table for radioactive isotope

    Isotope Half-life(years) Initial quantity Amt after 1000 years

    5715 ? 3.5g

    3. The population of P of a city is given by P= 105300 e^0.015t, where t represents the year with t=0 corresponding to 2000. Sketch the graph of this equation.According to this model, in which year will city have population of 150,000?
    4. The management at a factory has found that the maximum number of units a worker can produce in a day is 30. the leaning curve for the number of units N produced per day after a new employee has worked t days is given by

    After 20 days, a worker produced 19 units in 1 day.
    a. Find the learning curve for this worker (first hand the value of k)
    b. How many days should pass before this worker is producing 25 units/day?

    5. Plot the complex number : 1 - 2i

    6. Evaluate expression. Round 3 places

    7. Sketch graph of rational function as sketching aids, check for intercepts, symmetry, vertical asymptotes, and horizontal asymptotes

    i. ii.

    8. approximal real zeros with zoom and trace

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    https://brainmass.com/math/basic-algebra/logarithms-exponents-applications-word-problems-88311

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    1. Complete table for savings in which interest is compounded continously.

    Initial investment Annual rate Time to double total
    $1000 ? ? $2281.88

    Solution: There is missing information in this problem. The formula for continuous compound interest is A = P exp(rt) where P is the principal, r is the annual interest rate, t is the time in years, and A is the amount after time t. All we know is A = $2281.88 and P = $1000; so exp(rt) = P/A or rt = ln(P/A).

    If we knew r, we could solve for the time to double, d, by solving the equation
    2P = A. Since exp(rt) = P/A, this means exp(rt) = 1/2 , or rt = ln(1/2), so d = t = ln(1/2)/r.

    2. Complete table for radioactive isotope

    Isotope Half-life(years) Initial quantity Amt after 1000 years

    5715 ? 3.5g

    Solution:
    Radioactive substances decay exponentially. That is, the mass M after time t is given by the equation M = I exp(rt) , where r is a decay rate and negative, and I is the initial amount. The half-life ...

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