regarding ^x

[prog_tom]

Julie has $100 bucks,
she wants to buy stuff value of $999992
the value of that drops 12% every hour,
how many hours does she need to wait to buy that thing with her $100?

[Guv]

100 = 999992(.88)^x

Solve above for x

Using my HP calculator, I got 72 hours, if she has some change in her purse as well as the $100.00

[wy125]

Guv's right.

You can solve this equation directly by doing the following:

You started with an equation of the form

P = Q*r^x


In your case, P = 100, Q = 999992, and r = 0.88.

Divide both sides by Q to give:

P/Q = r^x

Now take the log of both sides so that you have

ln(P/Q) = x ln(r)

solving for x gives,

x = ln(P/Q)/ln(r)

x = ln(100/999992)/ln(0.88)

x = 72.049...

[prog_tom]

So, after 72 hours, the value of that thing, $999992, dropping hourly at rate of 88% will be $100?

[prog_tom]

BTW what is HP Calculator? Is it a peculiar one?

[prog_tom]

what do you do when the log is in base of e? I mean I don't want to use a calculator, is that possible? Also,
how did you switch In(P/Q) with x?
ln(P/Q) = x ln(r)
to
x = ln(P/Q)/ln(r)

Thanks

[Guv]

what do you do when the log is in base of e? I mean I don't want to use a calculator, is that possible? Also,
how did you switch In(P/Q) with x?
ln(P/Q) = x ln(r)
to
x = ln(P/Q)/ln(r)

Divide both sides by ln(r), then switch left & right sides.

Base ten logs can be used instead of base e.

HP stands for hewlitt Packard. The HP48 series were (might still be) as good as any other calculator and better than most. Mine might be slightly obsolete now.

It is programmable, providing a very sophisticated language.

It uses a stack instead of registers and algebraic input, which is the best architecture for either a computer or a calculator, although few computers use that architecture any more.

[prog_tom]

by ten logs you mean:

log10 x
----------
log10 a

?????????

BTW< HOw much do those caculators cost?

[JungleMan]

Originally posted by prog_tom
by ten logs you mean:

log10 x
----------
log10 a

?????????

BTW< HOw much do those caculators cost?

Check on Dealtime or Pricewatch..

If you really want a high end calc the Texas Instruments ones are good...I've got a Ti-83 plus silver, isn't exactly high end but I don't need high end I'm only in alg 2

[NotLKH]

If you really want a good one, upgrade from an HP to a PHP!

[Guv]

NotLKH: What is a PHP?

I am familiar with TI calculators and consider then inferior to HP.

[NotLKH]

Sorry, Guv,

Just messing about, There is no such thing as a PHP calc.

And you're quite right, TI Is inferior to HP,
BUT, I remember in the good old days, TI's were much cheaper than HP, so my first Advanced calc was a TI.

3 months later, I got an HP, Programmable, Pop in Modules...
Very Nice.

[prog_tom]

I found out another way:

100e^0.12x=999992
x=76.7525

76 hours 45 minutes and 9 seconds

Which way would you prefer?

[Guv]

Prog_Tom: I prefer to solve the equation which results in the correct answer.

72 hours is correct if you have about a nickel in change to go with the $100.00

Either your equation is wrong or you did not solve it correctly. I think it is the equation which is wrong.

The above might seem nasty, but I meant it to be humorous, not a put down.

[prog_tom]

1000=9999(0.12)^x

x= 1 hour and something

is that true?

A=Pe^rt is a formula given in Alg 2 book. e=2.718281828459; that's how I found my answer:
100e^0.12x=999992
x=76.7525

[Guv]

The original problem as stated has nothing to do with e.

The price drops 12% or .12 per hour.

The above means that every hour is is worth 88% or .88 times what it was worth the previous hour. The first few hours result in the following.

1 Hour: 879,992.96
2 hour: 774,393.80
3 Hour: 681,466.55

The above obtained by multiply by .88 each hour. If you keep it up for 72 hours, the result is a little more than $100.00

The formulaee you are using are not applicable to the problem.

The only time I remember seeing e involved in interest calculations, it was due to continous compounding of interest.

Not yearly compounding, not monthly, not daily, not each second, but continuous compound interest (I think) uses a formula involving e.

This problem is a straight forward, compund discount problem.

There are at least two posts which give the correct answer. My original post does not show the algbraic manipulation. Another post does show it.

Any type of logarithms could be used. My HP calculator provides both. Some calculators and compilers only provide one or the other.

[prog_tom]

Another question,

what is continuosly compound, when does it apply?

A girl has a newspaper biz, 4 new subscribers each day after her first day, how many subscribers will she have on her 25th delivery day?

[prog_tom]

how long will it take 9999 at rate of 12%/hour to drop to 1000?

1000=9999(1-0.12)^x


x=18 hours?

a guy from Harvard says it's wrong.

[Guv]

By my calculations, you need about a penny more than $1000 after 18 hours, so that looks like a good answer to me.

[Guv]

Prog_Tom: You asked about continuous compounding of interest.

It is a hypothetical concept. As far as I know, it is never done in practice. The idea is as follows.

The amount at the end of one year is (1 + r) if simple interest is applied to a $1.00 amount.

6% per year (for example) results in the following.

Simple interest: (1 + .06) = 1.06

If compounded monthly: (1 + .06/12)^12 = 1.061 678

If compounded daily: (1 + .06/365)^365 = 1.061 831

If compounded hourly: (1 + .06/8760)^8760 = 1.061 836 374

Note: e^.06 = 1.061 836 547

Note that as the compounding period gets smaller, the effective rate approaches e^r for one year.

Continuous compounding means that the compounding period approaches zero, in which case, the effective rate is e^r