Question

Can somebody help me? I am learning about Basic Exponent Properties in Algebra and I do not understand how to solve equations with exponents. Please tell me how to do it?

Here are some examples:
1. (4xy²)(-2x²y³)
2. 6y³/18x · 2xy
3. Which expression has the greater value? 2³ · 2(exponent 5) or 4(exponent 7)/4³
4. (3x)exponent1/3 · (3x)exponent7/3 / (3x)exponent2/3
Please Answer ASAP!! I will give 40 points!!

Answer 1

Provided an explanation below.

The examples are not asking you to solve, only to simplify the expression

For the examples you posted, not sure if you want the solutions are not; I have gone ahead and simplified

`1.\; -8x^3y^5\\\\2\; (2y^3)/(3)\\\\3.\; They are both equal\\\\4.\; 3x^2\\\\`

Explanation:

First let's understand what an exponent is.

When a number or a term or an expression is multiplied repeatedly by itself, an exponent can be used to represent this situation

for example,

5 x 5 x 5 x 5 x 5 x 5 = 5⁶ since 5 is multiplied by itself 6 times

Here the number being multiplied(5) is called the base and the number of times it is multiplied(6) is called the exponent.

There are some rules regarding exponentiation

1 Zero Exponent

Any number raised to the power zero is 1

`4^0 = 1\\\\123.456^0 = 1\\\\x^0 = 1`

2. Negative Exponent
If the exponent if negative, the expression is equivalent to the reciprocal of the term raised to the positive exponent

In general

`a^(-m) = (1)/(a^m)\\\\and\\\\(1)/(a^(-m)) = a^m\\\\`

Examples:

`x^-1 = (1)/(x)\\\\\left({x + y}\right)^(-a) = (1)/((x+y)^a)`

`(1)/(x^(-3)) = x^3`

3. Product Rule
When you multiply exponents with the same base the exponents get added

In general

`\displaystyle {a^m a^n = a^(m + n)}`

Examples

`x^3x^6 = x^((3+6)) = x^9\\\\`

4. Quotient Rule

When you divide an exponentiated term by another exponentiated term with the same base the resulting exponent is the difference between the two exponents:

In general

`(a^m)/(a^n) = a^{m-n`

Examples

`(y^3)/(y^2) = y^(3-2) = y^1 = y`

5. Power Rule 1

An exponent term raised to another power will be the base raised to the product of the two exponents

In general

`(a^m)^n = a^(mn)`

`(x^3)^4 = x^(3 \cdot 4) = x^(12)`

6. Power Rule 2

If the base is the product of two or more terms and the whole expression is raised to a power then each individual term gets raised to that power.

`(ab)^m = a^mb^m\\\\(2x^3y^2z^4)^4 = 2^4x^(3.4)y^(2.4)z^(4.4) = 16x^(12)y^8z^(16)`

7. One exponent
Any term raised to the power 1 is itself

`a^1 = a`

example: 5¹ = 5

x¹ = x

You may have to use some or all of the rules when confronted with a specific problem

There are plenty of excellent resources on the web which can explain far more lucidly than I can.

Here are a few. Just search for them and you will get the site links
LibreTexts, OpenStax, cK-12.org etc

As for the specific examples you posted:

`1. (4xy^2)(-2x^2y3)\\`

• Multiply the constant coefficients: 4 x -2 = -8
• Multiply each term with the same base by the other term with the same base
  
  `x\cdot x^2 =x^1x^2 = x^(1 + 2) = x^3\\\\\\y^2 y3 = y {2 + 3} = y^5\\\\`
  
• Combine all these to get the final answer:
  
  
  `(4xy^2)(-2x^2y^3) = -8x^3y^5`
  

`\textrm{2. }(6y^2)/(18x)\cdot 2xy`

• 
  `\mathrm{Cancel\:}(6y^2)/(18x):\quad (y^2)/(3x)\\\\`
  
• Then the expression becomes
  
  `=(y^2)/(3x)\cdot \:2xy`
• 
  `\mathrm{Apply\:the\:fraction\:rule}:\quad \:a\cdot (b)/(c)=(a\cdot \:b)/(c)`
  
  
  `(y^2)/(3x)\cdot \:2xy = =(y^2\cdot \:2xy)/(3x)`
  
• Cancel the x above and the x below to get
  
  
  `(y^2\cdot \:2y)/(3)`
  
• 
  `\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^(b+c)`
  
  `y^2y=y^(2+1) = y^3`
  
• Putting all the terms together we get
  
  `\textrm{2. }(6y^2)/(18x)\cdot 2xy =(2y^3)/(3)`
  
  

3. Which expression has the greater value?

`2^3.2^5 \; or\; (4^7)/(4^3)`

`2^3\cdot2^5 = 2^8\\\\(4^7)/(4^3)= 4^(7-3)=4^4\\\\\mathrm{Since \;4 = 2^2, 4^4 = (2^2)^4=2^8}`

`4. \frac{\left(3x^{(1)/(3)}\cdot 3x^{(7)/(3)}\right)}{3x^{(2)/(3)}}`

• 
  `\\\\\\\mathrm{Cancel\:the\:common\:factor:}\:3`
  ==>
  `\frac{3x^{(1)/(3)}x^{(7)/(3)}}{x^{(2)/(3)}}`
  
• 
  `\mathrm{Simplify\:}\frac{x^{(1)/(3)}}{x^{(2)/(3)}}`
  
  
  `\longrightarrow \frac{x^{(1)/(3)}}{x^{(2)/(3)}} = x^{(1)/(3) - (2)/(3) }= x^{-(1)/(3)} = (1)/(x^(1)/(3))\\\\\\\longrightarrow \frac{3x^{(7)/(3)}}{x^{(1)/(3)}} =3x^{(7)/(3)-(1)/(3)} = 3x^2`