I need someone fluent in math

[Herrlster]

to help me out with two questions. I'm not looking for any "do your own homework" type of comments because that's not what I'm asking for. I just need the steps, not the answer.

I'm taking gr.11 uni math right now and I need help with these questions.


1) Inflation causes things to cost roughly 2% more per year,

a) a movie ticket costs $8.50 now, when will the ticket cost $10.00?


b) how long ago did the ticket cost $4.25?


2) an element decays at a rate of 12% an hour, initially it was 100 g. when will there be 40 g left?

We're doing exponents and such. So the solution needs exponents.

[Lalabadie]

Those are exponential equations. You need to put the situation like this:

Every given time period, your initial value multiplies by a number (for example, if it's 2%, then it multiplies by 1.02 every time). To get the value after a number of years/days/seconds/etc, you need to use what you know in an equation like this:

x = y^n + i (I'm not using the standard variables, I don't remember them right now)

Where x is the number you're looking for, y is the initial value (i.e. 8.50 $), n is the multiplier (1.02 for a gain of 2%) and i an initial value that's not taken into account by the multiplier, if needed.

Replace the variables you know to get the one you want.

[Herrlster]

Yeah, they are exponential equations. Like, I know the basics like if the question asked how much it'd be worth in 10 years, its just

P=8.50(1.02)^10

But I don't know what to do when it gives me a value and I gotta figure out the time it took.

I think I need to do...

10=8.50(1.07)^1

but I don't know if thats right, or where to go after that...

[*smb]

I was taught to take the logs. eg x=y^z would become log(x)=log(y^z)=z*log(y)

[Herrlster]

Here's an example of another questions thats basically the same.

Bacteria doubles every 4 hours. Initially, there are 500 bacteria. How long will it take the bacteria to reach 4000.


P=4000
Po=500
Growth Rate=2
Growth Period=^t/4

4000=500 x 2^t/4

Divide the 500 out of the right side. So we need to divide 500 out of the left side aswell.

8=2^t/4

Now we need to make the bases the same.

2^3=2^t/4

Don't need to worry about the bases now.

^3=^t/4

Get rid of the 4 in the denom. by multiplying by 4. Need to multiply 3 by 4 aswell.

^12=^t

So it took 12 hours to reach 4000.


Now...can someone do that, but with my question? I get as far as here.

10=8.50 x 1.02^t/1

Divide both sides by 8.50

1.17647...=1.02^t/1

I cannot figure out how to get the same base on the left side. Any tricks or anything?

[IconBasser]

heh heh... I took this same chapter 2 months ago. I aced it!

[Geoff St. Germaine]

I'm not sure why you are putting the t/1 in, but here's how you go about it.

1.176... = 1.02^x

Using logarithms, we can write (taking _ to denote base):

log_1.02(1.02^x) = log_1.02(1.176...)

so,

x = log_1.02(1.176...)

A property of logarithms (that you may or may not have to prove, but I'm not doing it here) is:

log_b(c) = log(b)/log(c)

so, in this case:

x = log(1.176...)/log(1.02)

x = 8.206...

The other one can be solved the same way. I wouldn't use the t/4 though... I'd use t and multiply by 4 at the end.

[ManuelJG]

here's another property of logarithms that may help you:

Ln(y^x) = xLn(y)

in this case:

Ln(1.176) = xLn(1.02)

which brings the same answer that Geoff gave you, just without messing with the base of the logarithm:

x = Ln(1.176)/Ln(1.02)

[Fire-Starter]

All of you guys who have chimed in to help him, I have nothing but RESPECT for you. I have seen some of you in other threads helping out with Math and Physics, etc, including some of mine that I have posted. Great job guys. I wish I knew as much as some of you do in the Math area.

[TrooperFarva]

This is pretty basic stuff, it seems to have been covered, so nothing to add really.

[heroincredible]

Originally Posted by IconBasser

heh heh... I took this same chapter 2 months ago. I aced it!

Constructive post on your part.

[Disraeli Gears]

We must be in a similar math class, I had a test on exactly this today..

Anyway, for the first one, you're going to use the Compound Interest formula:

A=P(1+r/n)^nt

A = New Amount ($10)
P = Principle (Original Amount, $8.50)
R = Rate (2%)
N = Number of times compounded per year (in this case 1, since it raises 2% per year)
T = Time to get from A to P

10 = 8.5(1+.02/1)^t

Basically, you've just gotta get that into logarithmic form, then.. I don't remember from there... Starting to think mebeh I should've paid attention in that class...
For the second part of that one, plug everything in again (with 4.25 in place of 10), then switch the + to a -.

For the second one, I don't remember the formula..

[Liko]

1) OK, this is an exponential problem, along the lines of continuously-compounded interest (Pert). Tickets increase in price (P) at a rate of 2% (r) per year (t). The trick is that the questions it is asking ask for t, not the result of the expression for a given P, r, and t. The equation for A would look like:

x=Pe^rt (basic form)
10 = 8.5e^(1.02)t (substitute knowns)

You now have to rearrange the formula to solve for t.

10/8.5 = e^(1.02)t (isolate exponent)
ln (10/8.5) = ln(e^1.02t) (natural log of both sides)
ln(10)-ln(8.5) = 1.02t(ln(e)) (log properties)

(ln(10)-ln(8.5))/1.02 = t (isolate t; ln(e) = 1)

Plug that into your calculator and you're golden. Your answer is expressed in years; if you need it in months, days etc. then multiply. The answers to 1(b) and to 2 are arrived at in a similar fashion, however for 1(b) you will be expecting time to be negative, and for 2 the rate is less than 1. It is NOT negative (remember your exponent properties; x^-n = 1/(x^n))

[Lorenzini]

Maf howts my head. I want my ba ba.