# How can I solve 0 1

### Before it begins

For power equations, you should be able to handle powers and roots well. Here are the most important things in the overview, then you can also solve power equations well.

### What is a potency?

If you multiply a number by yourself, you can use the product as a **power** write.

$$5*5*5*5=5^4$$

└──┬───┘

$$ 4 $$ - times the factor $$ 5 $$

Exponent or high number

$$ uarr $$

$$5^4=625$$

$$ darr $$ $$ darr $$

Base power value

You can also use fractional and decimal numbers as well as real numbers as a basis:

**$$ cdot $$** $$(2/5)^2=(2/5)*(2/5)=4/25$$ **$$ cdot $$** $$(-0,3)^3=(-0,3)*(-0,3)*(-0,3)=-0,027$$.

The exponent (number of factors) is one **natural number.**

The **power** $$ a ^ n $$ der **real** Number $$ a $$ and the **natural** Number $$ n $$ is the product $$ a * a *… * a $$ of $$ n $$ factors. The calculation of the $$ n $$ power of a number is called $$ a $$ **Potentiate**.

You can count on potencies!

You can multiply powers with the same base by adding the exponents.

Example: $$ 10 ^ 3 * 10 ^ 2 = 10 ^ (3 + 2) = 10 ^ 5 $$

### What is a root?

The extraction of roots is the reverse of exponentiation.

Which number "to the power of 4" is 625? For this you need the root:

- $$ root 4 (625) = 5 $$, because $$ 5 ^ 4 = 625 $$
- $$ root 3 (8) = 2 $$, because $$ 2 ^ 3 = 8 $$

The **Pulling roots** is the reverse of exponentiation.

Terms:

Root exponent

$$ uarr $$

$$ root 3 (8) = 2 $$ $$ rarr $$ root value

$$ darr $$

Radicand

The $$ n $$ - th root $$ root n (b) $$ der **positive** real number $$ b $$ and the **natural** Number $$ n $$ is that **positive** Number $$ a $$, for which $$ a ^ n = b $$ applies. The calculation of the $$ n $$ - th root of a number is called $$ a $$ **Square root** and is the reverse operation to exponentiate.

**1. The root value is always positive.**

It is also $$ (- 5) ^ 4 = 625 $$ and it could be $$ root 4 (625) = -5 $$.

But the rooting has to be done **clearly** otherwise there would be "pointless" calculations such as

$$ root 4 (625) + root 4 (625) = 5 + (-5) = 0 $$.

So $$ root 4 (625)! = - 5 $$!**2. The radicand is always positive (or $$ 0 $$)**

It is $$ (- 2) ^ 3 = -8 $$ and it could be $$ root 3 (-8) = - 2 $$.

But: You can also view roots as powers with fractions as exponents, e.g. B. $$ root 3 (8) = 8 ^ (1/3) $$

Thus the contradicting calculation would be possible:

$$ - 2 = root 3 (-8) = (- 8) ^ (1/3) = (- 8) ^ (2/6) $$ $$ = (- 8) ^ (2 * 1/6) = root 6 ((-8) ^ 2) = root 6 (64) = 2 $$

with $$ - 2! = 2 $$.

So: no negative radicals!

### Power equations

You are now ready to solve equations with powers.

Equations of the form $$ x ^ n = b $$ with natural numbers $$ n, n> = 1, $$ and real numbers $$ b $$ are called **Power equations**.

All real numbers $$ x $$ that satisfy the equation are **solutions** the power equation.

**example**: $$ x ^ 3 = 27 $$

The solution is $$ x = 3 $$, since $$ 3 ^ 3 = 27 $$.

Or written with transformation:

$$ x ^ 3 = 27 $$ | $$ root 3 () $$

$$ x = root 3 (27) = 3 $$

$$ x = 3 $$

Power equations have the form $$ x ^ n = b $$ with $$ n in NN $$ and $$ n> = 1 $$. All real numbers $$ a $$ with $$ a ^ n = b $$ are solutions of the power equation.

In power equations of the form $$ x ^ n = b $$ you have to determine the basis of a power for the given natural exponent $$ n $$ and the real power value $$ b $$.

For $$ n = 2 $$ you get simple quadratic equations.

*kapiert.de*can do more:

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### Solve power equations $$ x ^ n = b $$

This is done quickly for $$ x ^ n = 0 $$. Then there is only the solution $$ x = 0 $$ for all $$ n $$, because $$ 0 ^ n = 0 $$ for all natural numbers $$ n $$.

For $$ b! = 0 $$ you differentiate between power equations with **straight** and with **odd** Exponents.

### Power equations with even exponents

The power equation $$ x ^ n = b $$ with even $$ n $$ only has a solution if $$ b> = 0 $$, e.g. B. $$ x ^ 2! = - 4 $$ for all $$ x $$.**example 1**

Equation: $$ x ^ 4 = 81 $$

Square root on both sides: $$ root 4 (x ^ 4) = root 4 (81) rArr x = 3 $$

Solutions: $$ x_1 = 3 $$ and $$ x_2 = -3 $$, because $$ 3 ^ 4 = (- 3) ^ 4 = 81 $$

**Example 2**

Equation: $$ x ^ 4 = 56 $$

Square root on both sides: $$ root 4 (x ^ 4) = root 4 (56) rArr x = root 4 (56) $$

Solutions: $$ x_1 = root 4 (56) approx 2.74 $$ and $$ x_2 = -root 4 (56) approx -2.74 $$

Power equations $$ x ^ n = b $$ with **straight** natural numbers have $$ n $$ for $$ b in RR $$ and

**$$ b <0 $$**: no solution,

**$$ b = 0 $$**: a solution $$ x = 0 $$,

**$$ b> 0 $$**: two solutions $$ x_1 = root n (b) $$ and $$ x_2 = -root n (b) $$.

Powers with an even exponent are always positive.

For all $$ n in NN $$, $$ 0 ^ n = 0 $$.

The value of a root $$ root n (a) $$ is always positive.

### Power equations with odd exponents

The power equation $$ x ^ n = b $$ with **odd** $$ n $$ has one and only one solution for all real numbers $$ b $$.

**1st case**: $$ b> 0 $$**example**

$$ x ^ 3 = 125 $$ | $$ root 3 () $$ $$ rArr $$ $$ x = root 3 (125) = 5 $$

Solution: $$ x = 5 $$, because $$ 5 ^ 3 = 125 $$

**2nd case**: $$ b <0 $$**example**

$$ x ^ 3 = -64 $$

Subsidiary step: Solve equation with positive $$ b $$:

$$ x ^ 3 = 64 $$ | $$ root 3 () $$ $$ rArr $$ $$ x = root 3 (64) = 4 $$

Solution of the original equation: $$ x = $$ $$ 4 $$, because $$ (- 4) ^ 3 = (- 4) * (- 4) * (- 4) = - 64 $$.

Power equations $$ x ^ n = b $$ with **odd** natural numbers have $$ n $$ for all $$ b in RR $$ **a solution** and the solution for

**$$ b <0 $$**: $$ x = -root n (-b) $$,

**$$ b = 0 $$**: $$ x = 0 $$,

**$$ b> 0 $$**: $$ x = root n (b) $$.

For $$ b <0 $$ (2nd case) you cannot simply take the $$ n $$ - th root on both sides, since the root can only be drawn from non-negative numbers. The root $$ root n (b) $$ is not defined for $$ b <0 $$.

### "Extended" power equations

You can convert some equations into the form $$ x ^ n = b $$ by equivalent transformations.

**example**

$$ 2x ^ 3-4 = -10 $$

**1. Equivalent transformation**

$$ 2x ^ 3-4 = -10 $$ $$ | + 4 $$

$$ 2x ^ 3 = -6 $$ $$ |: 2 $$

$$ x ^ 3 = -3 $$

**2. Solve the power equation with $$ b <0 $$**

Subsidiary step: Solve equation with positive $$ b $$:

$$ x ^ 3 = 3 $$ | $$ root 3 () $$

$$ rArr x = root 3 (3) $$

Solution of the original equation:

$$ x = -root 3 (3) approx -1.44 $$

Always put “extended” power equations first in the form $$ x ^ n = b $$ and then solve them.

At **equivalent transformation** an equation does not change the solutions of the equation.

*kapiert.de*can do more:

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and tests - individual classwork trainer
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### Solve power equations graphically

For graphically solving power equations of the form

$$ x ^ n = b (b in RR $$ and $$ n in NN) $$

do you bring the graph of one**Power function**($$ f (x) = x ^ n $$) and one

**linear functions**($$ g (x) = b $$) to the cut.

**1. Power equations with even exponents**

Power equation: $$ x ^ 2 = 6.25 $$

Linear function: $$ g (x) = 6.25 $$

Power function: $$ f (x) = x ^ 2 $$

Intersections of the graphs: $$ S_1 (-2.5 | 6.25) $$ and $$ S_2 (2.5 | 6.25) $$.

Solutions to the power equation: $$ x_1 = -2.5 $$ and $$ x_2 = 2.5 $$

**2. Power equations with odd exponents**

Power equation: $$ x ^ 3 = -8 $$

Linear function: $$ g (x) = - 8 $$

Power function: $$ f (x) = x ^ 3 $$

Intersection of the graphs: $$ S (−2 | 8) $$

Solution of the power equation: $$ x = −2 $$

Power equations of the form $$ x ^ n = a $$ can be solved graphically by intersecting the graphs of the power function $$ f (x) = x ^ n $$ and the linear function $$ g (x) = b $$ . The $$ x $$ coordinates of the points of intersection are the solutions to the power equation.

The graph of the linear function $$ g (x) = b $$ is a straight line parallel to the $$ x $$ axis.

### And now in general

You can graphically see how many solutions have power equations $$ x ^ n = b $$ with even and odd exponents $$ n $$.

**1. Power equations with even exponents**

$$ f (x) = x ^ n $$ with $$ n $$ even

There are either none, one or two intersections. So none, one or 2 solutions to the power equation.

**2. Power equations with odd exponents**

$$ f (x) = x ^ n $$ with $$ n $$ odd

There is always an intersection of the power function with the straight line. So always a solution to the power equation.

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