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Thread: HELP! please. very urgent. precal.

  1. #1
    taylorr556
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    Exclamation HELP! please. very urgent. precal.

    ok here are some questions i need help with ASAP.

    1)express this using rational exponents: 5 times the square root of 81x^(3)y^(8) (theres also a little 4 above the beginning of the radical.)



    2)a scientist has 37 grams of a radioactive substance that decays exponentially. assuming k= -0.3, how many grams of radioactive substance remain after 9 days? round your answer to the nearest hundredth.

    3) solve using common logarithms: 4^(x-3)=7^x



    4)what interest rate is required for an investment with continuously compounded interest to double in 8 years?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by taylorr556 View Post
    ok here are some questions i need help with ASAP.

    1)express this using rational exponents: 5 times the square root of 81x^(3)y^(8) (theres also a little 4 above the beginning of the radical.)
    Hint: $\displaystyle 5 \sqrt[4] {81x^3y^8} = 5 \left( 81x^3y^8 \right)^{1/4}$

    see post #3 here to see how to change radicals to powers.

    now what do you think we should do?

    2)a scientist has 37 grams of a radioactive substance that decays exponentially. assuming k= -0.3, how many grams of radioactive substance remain after 9 days? round your answer to the nearest hundredth.
    the general equation for a problem like this is:

    $\displaystyle P(t) = P_0 e^{-rt}$ (the power should be negative for decay, in general we consider r or in this case, k, to be positive)

    here, $\displaystyle P(t)$ is the amount remaining after time t, $\displaystyle P_0$ is the initial amount, $\displaystyle r$ (or $\displaystyle k$) is the rate of decay, and t is time.

    given the information in your question, our equation for this particular problem is:

    $\displaystyle P(t) = 37e^{-0.3t}$

    you want $\displaystyle P(9)$, now continue

    3) solve using common logarithms: 4^(x-3)=7^x
    take log to the base 10 of both sides:

    $\displaystyle \log 4^{x - 3} = \log 7^x$

    $\displaystyle \Rightarrow (x - 3) \log 4 = x \log 7$ .........since $\displaystyle \log_a \left( x^n \right) = n \log_a x$

    now we simply solve for $\displaystyle x$, i leave that to you

    4)what interest rate is required for an investment with continuously compounded interest to double in 8 years?
    the equation we use for continuous compounding is the same one we use for exponential growth. the time it takes to double is 8 years, so we know the interest rate is $\displaystyle \frac {\ln 2}8$.

    we could have solved for this by solving $\displaystyle 2P_0 = P_0e^{8r}$ for $\displaystyle r$. but this formula works in general. that is $\displaystyle rt = \ln 2$ when $\displaystyle t$ is the half-life (in the case of radioactive decay) or $\displaystyle t$ is the doubling time (in the case of exponential growth)
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