15 Aug, 2024
· Mathematics

What value of n makes the equation true

(2x9yn)(4x2y10)=8x11y20(2x^9y^n)(4x^2 y^{10}) = 8x^{11}y^{20}

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Long Explanation

Explanation

Solving for n

To solve for the value of nn that makes the given equation true, we'll start by simplifying both sides of the equation. The given equation is:

(2x9yn)(4x2y10)=8x11y20(2x^9y^n)(4x^2 y^{10}) = 8x^{11}y^{20}

Step-by-Step Simplification:

First, let's simplify the left-hand side (LHS) of the equation.

Left-Hand Side Simplification:

(2x9yn)(4x2y10)(2x^9 y^n) (4x^2 y^{10})

Multiply the constants:

24=82 \cdot 4 = 8

Combine the powers of xx:

x9x2=x9+2=x11x^9 \cdot x^2 = x^{9+2} = x^{11}

Combine the powers of yy:

yny10=yn+10y^n \cdot y^{10} = y^{n+10}

So the left-hand side simplifies to:

8x11yn+108x^{11} y^{n+10}

Right-Hand Side:

The right-hand side (RHS) of the equation is already simplified as:

8x11y208x^{11} y^{20}

Equate Both Sides:

We can now equate the simplified forms of the LHS and RHS:

8x11yn+10=8x11y208x^{11} y^{n+10} = 8x^{11} y^{20}

Compare Terms:

Since the coefficients (8) and the terms with xx on both sides are identical, we focus on the yy terms to find the value of nn:

yn+10=y20y^{n+10} = y^{20}

Solution for n:

To solve for nn:

n+10=20n + 10 = 20

Subtract 10 from both sides:

n=10n = 10

Therefore, the value of nn that makes the equation true is 10.

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Emily Rosen

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Emily Rosen is a recent graduate with a Master's in Mathematics from the University of Otago. She has been tutoring math students and working as a part-time contract writer for the past three years. She is passionate about helping students overcome their fear of math through easily digestible content.

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Concept

Exponentiation Rules

Exponentiation Rules

Exponentiation is a mathematical operation involving two numbers, the base aa and the exponent nn. The result is denoted as ana^n and represents the base aa multiplied by itself nn times.

Basic Rules

  1. Product of Powers: When multiplying two powers with the same base, add the exponents.

    aman=am+na^m \cdot a^n = a^{m+n}
  2. Quotient of Powers: When dividing two powers with the same base, subtract the exponents.

    aman=amnfora0\frac{a^m}{a^n} = a^{m-n} \quad \text{for} \quad a \neq 0
  3. Power of a Power: When raising a power to another power, multiply the exponents.

    (am)n=amn\left( a^m \right)^n = a^{m \cdot n}
  4. Power of a Product: To raise a product to a power, apply the exponent to each factor.

    (ab)n=anbn(ab)^n = a^n b^n
  5. Power of a Quotient: To raise a quotient to a power, apply the exponent to both the numerator and the denominator.

    (ab)n=anbnforb0\left( \frac{a}{b} \right)^n = \frac{a^n}{b^n} \quad \text{for} \quad b \neq 0

Special Cases

  • Zero Exponent: Any non-zero base raised to the power of zero is 1.

    a0=1fora0a^0 = 1 \quad \text{for} \quad a \neq 0
  • Negative Exponent: A negative exponent represents the reciprocal of the base raised to the absolute value of the exponent.

    an=1anfora0a^{-n} = \frac{1}{a^n} \quad \text{for} \quad a \neq 0

Examples

  • Product of Powers:

    2324=23+4=27=1282^3 \cdot 2^4 = 2^{3+4} = 2^7 = 128
  • Quotient of Powers:

    5652=562=54=625\frac{5^6}{5^2} = 5^{6-2} = 5^4 = 625
  • Power of a Power:

    (32)3=323=36=729\left( 3^2 \right)^3 = 3^{2 \cdot 3} = 3^6 = 729

Understanding exponentiation rules is crucial for simplifying expressions and solving equations involving powers.

Concept

Variable And Constant Simplification

Explanation of Variable and Constant Simplification

Variable and constant simplification is a process typically used in algebra and calculus to reduce mathematical expressions to their simplest form. This simplification involves combining like terms, reducing fractions, and simplifying expressions involving variables and constants.

Simplifying Variables

When simplifying expressions with variables, the goal is to combine like terms, which are terms that have the same variables raised to the same powers. For instance, in the expression:

3x+5x2x3x + 5x - 2x

All the terms are "like terms" because they each contain the variable xx. We can combine these terms by adding or subtracting their coefficients:

(3+52)x=6x(3 + 5 - 2)x = 6x

Simplifying Constants

Constants are numbers without variables. Simplifying constants often involves straightforward arithmetic like addition, subtraction, multiplication, or division. For example:

8+43=98 + 4 - 3 = 9

Combining Variables and Constants

Sometimes, expressions involve both variables and constants. In these cases, simplify the variable terms and the constant terms separately, and then combine them. Consider the expression:

4x+3+2x54x + 3 + 2x - 5

First, combine the variable terms:

(4x+2x)+(35)(4x + 2x) + (3 - 5)

which simplifies to:

6x26x - 2

More Complex Expressions

For more complex expressions involving both variables and constants, and operations like multiplication or division, keep an eye on the order of operations (PEMDAS/BODMAS rules) and specific algebraic identities. Here’s an example:

2x(3x+4)5(x+1)2x(3x + 4) - 5(x + 1)

First, distribute the terms:

2x3x+2x45x512x \cdot 3x + 2x \cdot 4 - 5 \cdot x - 5 \cdot 1

This results in:

6x2+8x5x56x^2 + 8x - 5x - 5

Now, combine like terms:

6x2+3x56x^2 + 3x - 5

Example with Fractions

Simplifying expressions can also involve fractions. Suppose we have the equation:

2x+423x63\frac{2x + 4}{2} - \frac{3x - 6}{3}

First, simplify each fraction separately:

2x2+423x3+63\frac{2x}{2} + \frac{4}{2} - \frac{3x}{3} + \frac{6}{3}

This results in:

x+2x+2x + 2 - x + 2

Finally, combine the constants:

44

Key Points

  1. Combining Like Terms: Add or subtract like terms to simplify expressions with variables.
  2. Constant Simplification: Perform standard arithmetic operations on constants.
  3. Order of Operations: Use PEMDAS/BODMAS rules to handle complex expressions.
  4. Fractions: Simplify each fraction separately before combining them.

By following these steps, you can effectively simplify mathematical expressions involving variables and constants, making them easier to work with in algebraic and calculus problems.