Integer Exponents

Question

Please help me. My math professor gives us simple examples during lectures then complicated ones as our homework. It hurts. :(

 

Please kindly do it step by step. I may get some techniques. ^_^

 

1. 5^2 + 10^-1 over 5^-2 + 2^2

2. 4^-1 + 2^-3 over 5^0 - 4^-2

3. (10^1-n)^n over (1000^n)^(n + 1) multiply by 10,000^(n^2+2) over 100^(3-n)

4. 9^(2n-1) over (3^(n+1))^n multiply by (81^(n-1))^(n+1) over (27^n+2))^(n-1)

Answer 1

I think there's a typo in 3 so I'll do it with what I think is correct.

 

(10^(1-n)^n)/(1000^n)^n+1 is the left side which I'll do first

(10^(n-n^2))/[(10^3)^n]^(n+1)

[10^(n-n^2)]/[(10)^3n]^(n+1)

[10^(n-n^2)]/[10^(3n^2+3n)

when you divide you subtract exponents

n-n^2-3n^2-3n=-4n^2-2n

10^(-4n^2-2n) is the first factor

now we'll work on the second factor

(10,000)^(n^2+2)/(100)^(3-n)

(10^4)^(n^2+2)/(10^2)^(3-n)

(10)^(4n^2+8)/(10)^(6-2n)

again subtract exponents

4n^2+8-6+2n=4n^2+2n+2

(10)^(4n^2+2n+2) is the second factor

when you multiply you add exponents

(-4n^2-2n)+(4n^2+2n+2)=-4n^2-2n+4n^2+2n+2=2

10^2=100 is the simplest form

 

I'll look at 4 and see what I come up with. Later Jana.

I think there's a typo in 4 so I'll do it with the correction.

 

Here's #4. We'll do the left factor first, then the right factor, and then multiply the two factors.

 

9^(2n-1)/[3(n+1)^n]

(3^2)^(2n-1)/(3)^(n^2+n)

(3)^(4n-2)/(3)^(n^2+n)

(3)^(4n-2-n^2-n)

(3)^(-n^2+3n-2)   (first factor)

 

[(81)^(n-1)]^(n+1)/[(27)^(n+2)]^(n-1)

[(3^4)^(n-1)]^(n+1)/[(3^3)^(n+2)]^((n-1)

[(3)^(4n-4)]^(n+1)/[(3)^(3n+6)]^(n-1)

(4n-4)(n+1)=4n^2-4 and (3n+6)(n-1)=3n^2+3n-6

(3)^(4n^2-4)/3()^(3n^2+3n-6)

(3)^(4n^2-4-3n^2-3n+6)

(3)^(n^2-3n+2)  (second factor)

now multiply and add the exponents

(3)^(-n^2+3n-2+n^2-3n+2)

(3)^0=1  (anything raised to the zero power is 1 except 0^0 which does not exist in mathematics !!)

so the final answer to 4 is just 1

 

 

 

Answer 2

1.

(5² + 10^-1) / (5^-2 + 2²)             This is the only step that has to do with exponents.

(25 + 1/10) / (1/25 + 4)           Now everything is arithmetic. 

(251/10) / 101/25)                   To divide by a fraction, multiply by the inverse

(251/10) · (25/101)                  Reduce

(251/2) · (5/101)                      Multiply

1255/202

 

2. Is done similarly

 

3. and 4. Some typing issues here

 

(10^1-n)ⁿ / (1000ⁿ)^n-1 · 10000 ⁿ               I assume that the exponent should be n²+2, yet this 

program does not allow that. Am I corredt?

 

Answer 3

THX very much for trying to make the problems clear, but I'm not sure that I interpret them correctly.  So, let me give some rules and examples; you may be able to understand it and solve the problems  ◊⊃

 

Interger exponents may be positive, negative, or 0 (which is neither positive nor negative).

 

Any number x, except 0, to the zero power (x^0) = 1.

 

x^2 means x squared; that is, x*x

 

The rule of exponents is:

  With the same value, you may multiply the numbers and add the exponents:     x^3 * x^4 = x^7

  This makes sense:  xxx * xxxx = xxxxxxx

 

Now, if I haven't lost you yet, you are almost home free!

 

The difference with negative exponents is that they produce a value in the denominator:

                 x^-1 = 1/x

 

So, we probably want to rewrite any negative exponents as fractions with a positive exponent in the denominator.

And, if we have to add fractions, we will want to find a common denominator, as usual.

 

If there are multiple expressions, it really just clutters the page; it doesn't make the problem any harder (We just covered all the rules, right?)

 

For (1) and (2), there are actual numbers, so just be careful with the lines in fractions.  Make negative exponents positive when putting numbers in the denominator; calculate common denominators as needed;  and be very careful when cancelling anything (these problems sometimes trick confused students; so don't be one of those).

 

For (3) and (4), you keep n.  Since we mentioned adding exponents, I should say that:  (x^n)(x) = x^(n+1), but you probably knew that.

 

p.s., Extra parentheses, like ( ( ( these ) ) ), help to make expressions clear [ remember P-E-M-D-A-S ].