Question

(y ^ 4 * y ^ n)/(y ^ 2) = y ^ - 3 Find the value of n. (2 marks)

Answer 1

n = -5

Explanation:

Given:

`(y^4 \cdot y^n)/(y^2)=y^(-3)`

Multiply both sides by y²:

`\implies (y^4 \cdot y^n)/(\diagup \!\!\!\!\!y^2) \cdot \diagup \!\!\!\!\!y^2=y^(-3)\cdot y^2`

`\implies y^4 \cdot y^n=y^(-3) \cdot y^2`

`\textsf{Apply exponent rule} \quad a^b \cdot a^c=a^(b+c):`

`\implies y^(4+n)=y^(-3+2)`

`\implies y^(4+n)=y^(-1)`

`\textsf{Apply exponent rule} \quad a^(f(x))=a^(g(x)) \implies f(x)=g(x):`

`\implies 4+n=-1`

Solve for n by subtracting 4 from both sides:

`\implies 4+n-4=-1-4`

`\implies n=-5`

Answer 2

• n = - 5

===========

Given equation:

• 
  `(y^4*y^n)/(y^2) =y^(-3)`

Solve it for n:

• 
  `y^(4+n-2)=y^(-3)`
• 
  `y^(2+n)=y^(-3)`
• 
  `2+n=-3`
• 
  `n=-3-2`
• 
  `n=-5`

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Used properties:

• 
  `a^b*a^c=a^(b+c)` product of exponents with same base
• 
  `a^b/a^c=a^(b-c)` division of exponents with same base
• If
  `a^b=a^c` then
  `b = c` equal exponents with same base