Do you initially multiply the indices to get one common index or do you multiply each variable first and then break it down? I don't understand with multiple variables.

Question

Abigail Allen · Accepted Answer

You can only 'solve' if there is an equals sign in there some where.
Solve implies assign value to or to establish a ratio.

However, I have demonstrated some manipulation.

To make it more straight forward lets convert the indices to fractional form. #color(blue)("Introduction to the approach")# By example: Suppose we had #root(3)(2)# this is the same as #2^(1/3)# Suppose we hade #root(3)(2^5)# this is the same as #2^(5/3)# Suppose we hade #2^(1/3)xx2^(5/3)# this is the same as #2^(1/2+5/2) = 2^(6/3) = 2^2# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Answering the question")# Given : # root(3)(x^2y) color(white)("d")( sqrt(xy)-root(5)(xy^3)color(white)("d"))# #color(brown)("Using the above approach")# Write as: #color(red)(x^(2/3)y^(1/3))color(white)("d")(color(green)(color(white)("d")x^(1/2)y^(1/2) color(white)("ddd")-color(white)("ddd")x^(1/5)y^(3/5)color(white)("d")))# #(x^(color(red)(2/3)color(green)(+1/2))xxy^(color(red)(1/3)color(green)(+1/2)))color(white)("d") - color(white)("d")( x^(color(red)(2/3)color(green)(+1/5))xxy^(color(red)(1/3)color(green)(+3/5)) )# #color(white)("dd")(x^(7/6)xxy^(5/6)) color(white)("dd.d")-color(white)("ddd")(x^(13/15)xxy^(14/15) )" ".........Expression(1)# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Lets do a quick check on equivalents Known that #1/2 = (1xx2)/(2xx2)=2/4# So is the following also true: #3^(1/2) -> 3^(2/4)# ? #sqrt(3)/root(4)(9) = 1# as required so we can do this. Note that #7/6 -> (7xx5)/(6xx5) =35/30# #5/6->(5xx5)/(6xx5)=25/30# #13/15->(13xx2)/(15xx2)=26/30# #14/15->(14xx2)/(15xx2)=28/30# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ So #Expression(1)# becomes #color(white)("dd")(x^(35/30)xxy^(25/30)) color(white)("dd.d")-color(white)("ddd")(x^(26/30)xxy^(28/30) )" ".........Expression(1_a)# From this we can factor out the least values #x^(26/30)y^(25/30)color(white)("d")(color(white)("d")x^(9/30)-y^(3/30))# #x^(13/15)y^(5/6)( x^(3/10)-y^(1/10))# Horrible numbers!!!!!!

Abigail Allen · Answer

undefined When dealing with multiple variables raised to exponents, it's essential to understand the rules of exponents and how they apply to each term.

Let's clarify the process:

1. **Multiplying Exponents with the Same Base**:
   When you have variables with the same base raised to different powers, you multiply the exponents to combine them.

Example: $ x^a \times x^b = x^{a+b} $

2. **Multiplying Multiple Variables Raised to Exponents**:
   If you have multiple variables raised to different exponents being multiplied together, you can apply the exponent rule for each variable separately.

Example: $ (x^a)(y^b) $
   - Multiply the exponents for $ x $ and $ y $ separately:
     $ x^a \times y^b = x^a \times y^b $

Here's a step-by-step breakdown when you have an expression with multiple variables raised to exponents being multiplied together:

1. Multiply the coefficients (if they are not 1).
2. Apply the exponent rule for each variable separately.
3. Combine the results.

Example:

Given $ (x^2y^3)(x^4y^2) $

1. Multiply the coefficients (in this case, 1 for both):
   $ 1 \times 1 = 1 $

2. Apply the exponent rule for $ x $ and $ y $ separately:
   $ x^2 \times x^4 = x^{2+4} = x^6 $
   $ y^3 \times y^2 = y^{3+2} = y^5 $

3. Combine the results:
   $ (x^2y^3)(x^4y^2) = x^6y^5 $

Always remember to simplify each term separately and then combine them.