I'm new to this. Help?

Master Ryu

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Aug 9, 2005
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I didn't really see any notes about this problem. Simplify (-9) to the 2nd power. I also need help with evaluating (5.5) with an exponent of -2. Also,
is (0.5) to the third power divided by (0.5) to the sixth power (0.5) to the third power? I still need to finish the rest of my homework and I'll edit my post with some of the other tproblems I don't understand.
 
Master Ryu said:
Simplify (-9) to the 2nd power.
Squaring just means "times itself", so, for instance, 3<sup>2</sup> = (3)(3) = 9. So multiply -9 by itself.

Master Ryu said:
I also need help with evaluating (5.5) with an exponent of -2.
Multiply 5.5 by itself.

Master Ryu said:
Also, is (0.5) to the third power divided by (0.5) to the sixth power (0.5) to the third power?
No. You do have three extra copies of 0.5, but they are in the denominator, not the numerator.

. . . . .(0.5<sup>3</sup>)/(0.5<sup>6</sup>) = (0.5×0.5×0.5)/(0.5×0.5×0.5×0.5×0.5×0.5) = 1/(0.5×0.5×0.5)

Simplify.

Eliz.
 
Oops, I'm sorry, but the exponents were outside the ( ). So, am I still incorrect? I wish my teacher was this clear.

EDIT : Also, I need help with this problem that makes no sense to me.

Find the largest value of n such that 3 with exponent -n > 0.00001.
 
Inside or outside, if it is just a number it doesn't matter.
(6²) = (6)²
It is asking how big n can be and still have
(1/3)^n > .00001
If you can use logs
n*log(1/3) > log(.00001) = -5
or
n = -5/log(1/3)
drop the decimals for the answer.
otherwise keep multiplying
(1/3)*(1/3)*(1/3)...
til it is smaller than .00001. The answer is one less than the number of (1/3)s you used
 
Master Ryu said:
[Oops, I'm sorry, but the exponents were outside the ( ). So, am I still incorrect? I wish my teacher was this clear.]

use symbol ^ instead of "with exponent";
.5^3 is SAME as (.5)^3, so don't get all excited :)

[EDIT : Also, I need help with this problem that makes no sense to me.
Find the largest value of n such that 3 with exponent -n > 0.00001.]

That's 3^(-n) > .00001
RULE: a^(-x) = 1 / a^x
so: 1 / 3^n > .00001

if n=10, then 1 / 3^10 = 1 / 59049 = .0000169...
if n=11, then 1 / 3^11 = 1 / 177147 = .00000564...
So largest value of n is 10.

I used trial and error; don't think you're ready for logs yet;
but if you are, here's how:
1 / 3^n = .00001
3^n = 1 / .00001
3^n = 100000
n = log(100000) / log(3)
n = 10.479...
So largest value of n (if integer) is 10
Edit: Gene, you're making me nervous!
Btw, am I correct or wrong?
 
I just kinda got confused since my teacher told me to subtract exponents when dividing and the base number is the same.

And thanks so much with that tough question. First day learning some of this. I'm in advanced classes and no doubt I'll be learning that stuff next class when we turn in our homework.
 
Master Ryu said:
...my teacher told me to subtract exponents...
That's fine, but do it sensibly. If the exponent is larger underneath, then that's where you have the extra copies. Subtract, but leave the left-overs underneath.

. . . . .x<sup>5</sup>/x<sup>2</sup> = x<sup>5-2</sup> = x<sup>3</sup>

. . . . .x<sup>2</sup>/x<sup>5</sup> = 1/x<sup>5-2</sup> = 1/x<sup>3</sup>

Eliz.
 
Denis,
Since (I think) we said exactly the same thing I don't understand your nerve attack.
---------------
Gene
 
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