exponent/raised to a power on calculator

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glme

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for the compound interest questions, sometimes I would have to do raised to the 12th power, for which I would prefer to use a calculator, but I don't think the calculator on BC, which should be identical to the actual DAT calculator, has one?

If it really does not have one, how do you solve 0.01^12?

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for the compound interest questions, sometimes I would have to do raised to the 12th power, for which I would prefer to use a calculator, but I don't think the calculator on BC, which should be identical to the actual DAT calculator, has one?

If it really does not have one, how do you solve 0.01^12?
Here's the right answer:
(.01)^12 = (10^(-2))^12 = (10)^(-24)
Note:
0.01 = 10^(-2)
 
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The DAT calculator really doesn't have a power function. The expectation here is that you will learn the methods above to manipulate exponents manually rather than solving them out manually.
 
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The DAT calculator really doesn't have a power function. The expectation here is that you will learn the methods above to manipulate exponents manually rather than solving them out manually.

Is it the same one used for gChem? How are we supposed to find pH/[H3O+] with acid base problems if there's no log/exponent functions?
 
Here's the right answer:
(.01)^12 = (10^(-2))^12 = (10)^(-24)
Note:
0.01 = 10^(-2)


I understand how to get to that, but here's the question he is referring to:


Craig invests $1,000 in two different banks. At Bank A, Craig deposits his money into a new type of account that has a monthly interest rate of 1% (rather than an annual rate) and compounds every month. At Bank B, he invests in a normal account with an annual interest rate of 12% compounded annually. Which investment made more money at the end of one year, and by how much?

Bank B is easy to solve for, but for bank A, you end up at:

math-5-38-2.png



How do you solve for that exponent? Do I have to plug it into my calculator 12 times?

Thanks
 
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Is it the same one used for gChem? How are we supposed to find pH/[H3O+] with acid base problems if there's no log/exponent functions?

You should be able to approximate. For example if [H+] = 1 * 10^-7, then you can say the -log = 7. Similarly, 1*10^-8 : pH =8. Also, know the half pH values which occurs at [H+] = 3.16 * 10^whatever. If you have Chad's videos, he does a great explanation.
 
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