Math Problem Statement

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Solution

Let's go through the exercises from the document. Here's the solution for each problem:

1. Schreibe mit nur positiven Exponenten (Write with only positive exponents):

  • 5x1=5x5x^{-1} = \frac{5}{x}
  • 23=123=182^{-3} = \frac{1}{2^3} = \frac{1}{8}
  • 32x1=9x3^2 x^{-1} = \frac{9}{x}
  • (2x)3=1(2x)3=18x3(2x)^{-3} = \frac{1}{(2x)^3} = \frac{1}{8x^3}
  • a14=1a14=1a4a^{- \frac{1}{4}} = \frac{1}{a^{\frac{1}{4}}} = \frac{1}{\sqrt[4]{a}}
  • (ab)1=ba\left(\frac{a}{b}\right)^{-1} = \frac{b}{a}

2. Vereinfache mit Hilfe der Potenzgesetze (Simplify using power laws):

  • x3x4=x3+4=x7x^3 \cdot x^4 = x^{3+4} = x^7
  • a5a2=a52=a3a^5 \cdot a^{-2} = a^{5-2} = a^3
  • z3z4z1=z3+41=z6z^3 \cdot z^4 \cdot z^{-1} = z^{3+4-1} = z^6
  • (a2b3)(4a2b)=4a2+2b3+1=4a4b4(a^2 b^3) \cdot (4a^2 b) = 4a^{2+2} b^{3+1} = 4a^4 b^4
  • x7x5=x75=x2\frac{x^7}{x^5} = x^{7-5} = x^2
  • a2a6=a26=a4=1a4\frac{a^2}{a^6} = a^{2-6} = a^{-4} = \frac{1}{a^4}
  • a4a2=a4+2=a6\frac{a^4}{a^{-2}} = a^{4+2} = a^6
  • a3b2a4b=a34b211=a1b11=ba\frac{a^3 b^2}{a^4 b} = \frac{a^{3-4} b^{2-1}}{1} = \frac{a^{-1} b^1}{1} = \frac{b}{a}
  • (3x2)3=33x23=27x6(3x^2)^3 = 3^3 x^{2 \cdot 3} = 27x^6
  • (4a2)4=44a24=256a8=256a8(4a^{-2})^4 = 4^4 a^{-2 \cdot 4} = 256 a^{-8} = \frac{256}{a^8}
  • (aa2)=a1+2=a3(a a^2) = a^{1+2} = a^3
  • 4(2x)2=44x2=16x24 \cdot (2x)^2 = 4 \cdot 4x^2 = 16x^2

3. Vereinfache/Schreibe als eine Potenz (Simplify/Write as a single power):

  • a4b4=(ab)4a^4 b^4 = (ab)^4
  • a5a3=a53=a2\frac{a^5}{a^3} = a^{5-3} = a^2
  • (ab)3(ab)=(ab)3+1=(ab)4(ab)^3 \cdot (ab) = (ab)^{3+1} = (ab)^4
  • (2x)2(3y)2=4x29y2=36x2y2(2x)^2 \cdot (3y)^2 = 4x^2 \cdot 9y^2 = 36x^2 y^2

4. Vereinfache soweit wie möglich (Simplify as much as possible):

This part does not have explicit problems listed, but I can assume the task would involve simplifications like those in previous sections.

5. Schreibe als Wurzel bzw. Potenz (Write as a root or power):

  • x12=xx^{\frac{1}{2}} = \sqrt{x}
  • x13=x3x^{\frac{1}{3}} = \sqrt[3]{x}
  • 3215=325=232^{\frac{1}{5}} = \sqrt[5]{32} = 2
  • 24315=2435=3243^{\frac{1}{5}} = \sqrt[5]{243} = 3

6. Vereinfache (Simplify):

  • 2723=273=24=162^7 \cdot 2^{-3} = 2^{7-3} = 2^4 = 16
  • 31228=(33)428=27428=531441256=1362812163^{12} \cdot 2^8 = (3^3)^4 \cdot 2^8 = 27^4 \cdot 2^8 = 531441 \cdot 256 = 136281216
  • 8x232x=256x38x^2 \cdot 32x = 256x^3
  • x6y4x5y3=x65y431=xy\frac{x^6 y^4}{x^5 y^3} = \frac{x^{6-5} y^{4-3}}{1} = xy
  • (2a2b3)1a2b1=12a2b3a2b=b2\frac{(2a^2 b^{-3})^{-1}}{a^{-2} b^{-1}} = \frac{1}{2a^2 b^{-3}} \cdot a^2 b = \frac{b}{2}
  • 8x24y5x516y=32x2y5x5y=32x3y4\frac{8x^2 \cdot 4y^5}{x^5 \cdot 16y} = \frac{32x^2 y^5}{x^5 y} = \frac{32}{x^3 y^4}

These are the solutions for each problem listed in the document.

Would you like any further details or have any questions about these solutions?

Related Questions:

  1. How do you apply the laws of exponents when simplifying fractions?
  2. How do negative exponents impact the base and how can you rewrite them?
  3. What are the steps to convert a power into a root, like x1/2x^{1/2}?
  4. How do you simplify powers with multiple variables, such as (ambn)(a^m b^n)?
  5. What is the general rule for multiplying two expressions with the same base but different exponents?

Tip:

When working with exponents, always remember that multiplying powers with the same base means adding exponents, and dividing powers means subtracting exponents.

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Math Problem Analysis

Mathematical Concepts

Exponentiation
Roots
Power Laws
Negative Exponents

Formulas

a^m * a^n = a^(m+n)
a^m / a^n = a^(m-n)
(a^m)^n = a^(m*n)
a^(-n) = 1/a^n
x^(1/2) = √x

Theorems

Laws of Exponents
Simplification using power rules

Suitable Grade Level

Grades 9-11