# HELP! please. very urgent. precal.

• Jan 8th 2008, 04:17 PM
taylorr556
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?
• Jan 8th 2008, 04:37 PM
Jhevon
Quote:

Originally Posted by taylorr556
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: $5 \sqrt[4] {81x^3y^8} = 5 \left( 81x^3y^8 \right)^{1/4}$

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

now what do you think we should do?

Quote:

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:

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

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

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

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

you want $P(9)$, now continue

Quote:

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

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

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

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

Quote:

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 $\frac {\ln 2}8$.

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