# Thread: Just a Few Calculus Assignment Questions

1. ## Just a Few Calculus Assignment Questions

Alright, I'm back and looking for somemore help

1) I'm having some trouble finding the X intercept using logarithms for y=2(0.5)^X-6. Here are my steps so far:

log0= xlog(2(0.5))-6
log0+6/log(2(0.5))=x

This won't let you get an answer, I assume this is because log(1) is 0 because that is what my calculator is telling me. According to the graph I made with table of values, the x intercept should be arround -1.75.

Also, that question is asking for the equation of the asymptotes. Can anyone fill me into what that would be?

2) Express the function y=5^x as an equivilent function with base 3.

3) Suppose money can be invested at 3.75% compounded semi-annually. Determine how long it takes for a principle to triple in value. (I have an idea on how to do this up to a certain point but then I get stuck.)

4) The equation D= 10logI models the decible level of a sound whose intensity is I watts per sq. meter W/m2. The decible levels of a subway and normal conversation are 115dB an 60db respectivly. How many times as intense as normal conversation is the noise of a subway?

If anyone could help with any of these I'd be very greatful. This is a wonderful resource for the struggling calc. student.

2. Originally Posted by AndrewV
1) I'm having some trouble finding the X intercept using logarithms for y=2(0.5)^X-6. Here are my steps so far:

log0= xlog(2(0.5))-6
log0+6/log(2(0.5))=x

This won't let you get an answer, I assume this is because log(1) is 0 because that is what my calculator is telling me. According to the graph I made with table of values, the x intercept should be arround -1.75.

Also, that question is asking for the equation of the asymptotes. Can anyone fill me into what that would be?
Right idea, but you're doing the logs too soon.
You are looking for all points of the form (x, 0). So:
$0 = 2 \left ( \frac{1}{2} \right )^x - 6$

$6 = 2 \left ( \frac{1}{2} \right )^x$

$3 = \left ( \frac{1}{2} \right )^x$

$\frac{1}{3} = 2^{x}$

$x = log_2 \left ( \frac{1}{3} \right )$

As to asymptotes, we don't find vertical asymptotes for exponential functions. (Unless they are composite, as in the case of something more complicated like $f(x) = \frac{x - 3}{2^x - 1}$ or some such.) We do get horizontal asymptotes. Take the limit of your function as x goes either to positive or negative infinity:
$\lim_{x \to -\infty} (2(0.5)^x-6) = -6$.

-Dan

3. Originally Posted by AndrewV
Alright, I'm back and looking for somemore help

2) Express the function y=5^x as an equivilent function with base 3.

3) Suppose money can be invested at 3.75% compounded semi-annually. Determine how long it takes for a principle to triple in value. (I have an idea on how to do this up to a certain point but then I get stuck.)

4) The equation D= 10logI models the decible level of a sound whose intensity is I watts per sq. meter W/m2. The decible levels of a subway and normal conversation are 115dB an 60db respectivly. How many times as intense as normal conversation is the noise of a subway?

If anyone could help with any of these I'd be very greatful. This is a wonderful resource for the struggling calc. student.
2) y = 5^x.
You mean, in the form y = 2^z

y = y
5^x = 2^z
Take log of both sides,
x*log(5) = z*log(2)
So, z = x*log(5) /log(2)
Therefore,
y = 5^x is same as y = 2^[x*log(5) /log(2)] -----------answer.

-------------------------
3) Compound interest.
A = P(1+i)^n
where
A = amount after n compounding times---------3P here.
P = principal or initial amount
i = r/m
r = interest rate per year, in decimals -----------0.0375 here.
m = number of times compounded per year-------2 here.
n = number of times compounded in any time

So,
3P = P[1 +(0.0375)/2]^n
3 = (1.01875)^n
Take the common log of both sides,
log(3) = n*log(1.01875)
n = log(3) /log(1.01875)
n = 59.14 times compounded
At twice per year, 59.14 /2 = 29.57 years or 30 years.

Therfore, in 30 years, the principal will triple in value. ------answer.

------------------------------------------------------------
3)
D = 10*log(I)
where
D = decibel level. (I assume. First time I've encountered this formula.)
log = ln = natural log. (I assume too. If common log, then the answer is too big.)
I = intensity.

In subway, we find intensty Is:
115 = 10*ln(Is)
11.5 = ln(Is)
Is = e^11.5

In conversation, we find intensity Ic:
60 = 10*ln(Ic)
6 = ln(Ic)
Ic = e^6

So, Is /Ic = e^11.5 /e^6 = e^(11.5 -6) = e^5.5 = 244.7

Therefore, the intensity of the noise in a subway is about 245 times the intensity of the noise in a conversation. -----------answer.

4. Originally Posted by AndrewV
4) The equation D= 10logI models the decible level of a sound whose intensity is I watts per sq. meter W/m2. The decible levels of a subway and normal conversation are 115dB an 60db respectivly. How many times as intense as normal conversation is the noise of a subway?
The $dB$ difference between the two intensities is $115-60=55\ dB$.

Therefore the subway is $10^{55/10}$ times which is approximatly 316227 times.

That is a ratio of two power densities in $dB$ is the difference of the $dB$
power densities.

$10 \log_{10} (I_1/I_2)=10 \log_{10} (I_1)-10 \log_{10}(I_2)$

RonL