## Exponent Examples Math

As examples, consider the following: ( 1 ) 64 x = 16 ( 2 ) 3 x 2 – 5 = 81 x ( 3 ) 3 x 2 – 5 = 8 x ( 4 ) 2 x + x 2 = 2 If the unknown occurs occurs only as an exponent, then one speaks of a pure exponential equation (examples 1, 2 and 3).

## Solving by Exponent Comparison

If a pure exponential equation is to be solved, where only one base of the exponents occurs or different bases can be traced back to the same one, one can apply the power laws and find the unknown by comparing the exponents. In the above examples 1 and 2 this is the case.

• Example 1: 64x=1

Because 64 = 2 6 a n d 16 = 2 4 , the equation to be solved is equivalent to ( 2 6 ) x = 2 4 and according to the power laws to 2 6 x = 2 4 . The two exponents must be equal, so: 6 x = 4 ⇒ x = 2 3 The sample confirms this solution, because it is: 64 2 3 = 64 2 3 = 4096 3 = 16 ( 16 3 = 4096 )

• Example 2: 3×2- 5=81x

Again, because of 81 = 3, 4 equal bases can be produced. Thus the initial equation is equivalent to: 3 x 2 – 5 = 3 4 x The exponent comparison yields x 2 – 4 x = 5 and thus the quadratic equation x 2 – 4 x – 5 = 0. According to the solution formula one obtains x 1 = 5 u n d x 2 = – 1. The sample for x 1 yields: l i n k e S e i t e : 3 25 – 5 = 3 20 = 3 4 ⋅ 5 = 81 5 right side: 81 5 For x 2 yields: l i n k e S e i t e : 3 1 – 5 = 3 – 4 = 81 – 1 right side: 81 – 1 The sample thus confirms the correctness of both solutions.

## Solving by logarithm

In example 3 it would be difficult to produce equal bases for the existing exponents. Such exponential equations (of course also those like the previous ones) can be solved by logarithmizing both sides and then applying the logarithm laws. Here, one can choose any positive number a ( m i t a ≠ 1 ) as the base of the logarithms. Since the decadic and natural logarithms, i.e., the logarithms to bases 10 and e are tabulated or can be easily determined with a calculator, one will generally choose one of these bases.

• Example 3: 3×2- 5=8x

Logarithmizing gives: lg ( 3 x 2 – 5 ) = lg 8 x ( x 2 – 5 ) ⋅ lg 3 = x ⋅ lg 8 Calculating with rational approximations gives lg 8 ≈ 0.90309, lg 3 ≈ 0.47712 and lg 8 lg 3 ≈ 1.8928 . This gives the quadratic equation x 2 – 1.8928 x – 5 = 0 . According to the solution formula, we obtain as rational approximations: x 1 ≈ 3.3745 a n d x 2 ≈ – 1.4817 The sample for x 1 yields: l i n k e s e i t s : 3 3.3745 2 – 5 ≈ 3 6.38725 ≈ 1115.6 right-hand side: 8 3.3745 ≈ 1115.2 For x 2 one obtains: l i n k e s e i t s : 3 ( – 1.4817 ) 2 – 5 ≈ 3 – 2.80457 ≈ 0.045907 right-hand side: 8 – 1.4817 ≈ 0.045908 The sample, calculated with rational approximations using a calculator, seems to confirm the correctness of both solutions. The minor discrepancies are likely to result from rounding errors. However, in contrast to the previous example, absolute certainty is not given. In order to achieve this, extensive accuracy considerations would have to be made for the calculations carried out or approximate values should not be used.

## Graphical solving

If no pure exponential equations are to be solved, graphical solving is suitable under certain circumstances. Such a case is given in the example 4 mentioned above.

• Example 4: 2x+x2=2

From 2 x + x 2 = 2 one obtains by transforming 2 x = – x 2 + 2. If one now takes the corresponding functions y = f ( x ) = 2 x and y = g ( x ) = – x 2 + 2, then solving the equation is equivalent to finding the abscissas of the intersections of the two function images. From the graph, you can read x 1 = – 1.25 a n d x 2 = 0.6. The sample for x 1 yields: l i n k e side : 2 – 1.25 + ( – 1.25 ) 2 ≈ 0.420448 + 1.5625 ≈ 1.98right side: 2 For x 2 yields: left side:20.6+(0.6)2≈1.51572+0.36≈1.88right side:2 Power rules The following rules apply to calculating with powers. They are applied when simplifying calculations. Precedence rules

• Parentheses first
• Power calculation before point calculation
• Point calculation before dash calculation
BasicA power with the exponent 0 has the value 1. A power with the exponent 1 has the value of the power base.
a0 = 1; a1 = a 50 = 1; 51 = 5
Base and exponent equalAddition and subtraction: Factors belonging to the base are added or subtracted.
an + an = 2an 3an + 2an = 5an 5an – 3an = 2an 32 + 32 = 2 · 32 3a2 + 2a2 = 5a2 5a2 – 3a2 = 2a2 a2 + 5x4 + a2 – 3x4 = 2a2 + 2x4
Base equalMultiplication: The exponents are added.
am – an = am + n 42 · 43 = (4 · 4) · (4 · 4 · 4) = 4(2 + 3) = 45
Division: The exponents are subtracted (applies to m > n).
am : an = am – n
 45 : 43 = 4 · 4 · 4 · 4 · 4 = 4(5 – 3) = 42 4 · 4 · 4
Exponent equalMultiplication and division: The corresponding bases are multiplied or divided.
an – bn = (ab)n an : bn = (a : b)n 22 · 32 = (2 · 2) · (3 · 3) = (2 · 3) · (2 · 3) = (2 · 3)2 62 : 32 = (6 · 6) : (3 · 3) = 6 · 6 : 3 : 3 = (6 : 3) · (6 : 3) = (6 : 3)2
Power of the powerPower: The exponents are multiplied. The base remains unchanged.
(am)n = am – n (42)3 = (4 · 4) · (4 · 4) · (4 · 4) = 4(2 · 3) = 46

## base and exponent equal Addition – Subtraction

Task 1: Fill in the missing values.

 a) 3 – 23 + 2 · 23 = · = b) 32 + 4 · 32 = · = c) 8 – 32 – 2 · 32 = · = d) 5 – 42 – 42 = · = e) 10 – 22 + · 22 = · 22 = 48 f) 10 – 23 – · 23 = · 23 = 32

correct: 0wrong: 0 Task 2: Fill in the missing values.

 a) 3 – 23 + 2 · 23 = · b) 32 + 4 · 32 = · c) 8 – 32 – 2 · 32 = · d) 5 – 42 – 42 = · e) 10 – p2 + – p2 = – p2 f) 10 – q3 – – q3 = – q3

correct: 0wrong: 0 Task 3: Fill in the missing values.

 a) x2 + x2 = · b) a5 + 4 – a5 = · c) 6 – m3 – 2 – m3 = · d) 4 – y6 – 3 – y6 = e) 5 – z3 + – = 12 – z3 f) -3 – b2 + – = 5 – b2

Attempts: 0 Task 4: Fill in the missing values.

 a) 6 – p4 + 2 – p4 = · b) 6 – pq4 + 2 – pq4 = · c) 9 – x7 – 3 – x7 = · d) 9 – xy7 – 3 – xy7 = · e) 12 – ab5 + – = 14 – ab5 f) – 3 – ab2 = 5 – ab2

Attempts: 0 Task 5: Enter the missing values. correct: 0wrong: 0 Task 6: Fill in the missing values.

 a) 4x2 – 2x3 – 5x3 + 3x2 + 9x3= x + x3 b) 9a7 + a4 – 6a4 – 5a7 + 2a4= a – a4 c) 12y3 + 7y5 – 9y4 + 3y4 + 5y3= y3 + y – y4 d) 9b2 + b4 – 3b4 + 7b3 + b2 = 13b2 + 2b4 + b3

Attempts: 0 Task 7: Fill in the missing values.

 a) 5(a2 + b3) – 2a2 + 4b3 = a + b b) (x5 – y7)8 – 2(x5 – y7) = x – y c) 2u3 + 9(v3 – u3) + 5(u3 – v3)= u + v

Trials: 0

## Base equal Multiplication – Division

Task 8: Fill in the missing values.

 a) 22 · 23 = b) 4 – 42 · 412 = c) 78 : 76 = d) 64 · = 612 e) 87 : = 84 f) : 52 = 57

correct: 0wrong: 0 Task 9: Fill in the missing values.

 a) 22 · 23 = b) 4 – 42 · 412 = c) 78 : 76 = d) 64 · = 612 e) 87 : = 84 f) : 52 = 57

correct: 0wrong: 0 Task 10: Summarize the terms. correct: 0wrong: 0 Task 11: Summarize the terms.

 a) x2 – x2 – x2 = b) a1 – a2 – a3 = c) bm – bn = d) y5 : y3 = e) xm : xn = f) (-a)2m : (-a)m = ()

Trials: 0 Task 12: Fill in the missing exponents.

 a) 25 · 2 = 29 b) 7 – 73 = 75 c) 43 · 4 = 46 d) x5 – x = x7 e) y – y4 = y8 f) a3 – a = a11

Trials: 0

## Exponent equal Multiplication – Division

Task 13: Fill in the missing values.

 a) = b) = c) – = d) = e) – f) )

correct: 0wrong: 0 Task 14: Fill in the missing values.

 a) = b) = c) : = d) = e) : f) )

correct: 0wrong: 0 Task 15: Fill in the missing values.

 a) 62 : 32 = 2 b) 167 : 27 = c) 125 : = 45 d) 186 : 4,56 = 6 e) 103 : = 43 f) from4 : b4 =

Try: 0 Task 16: Complete the simplified terms correctly. correct: 0wrong: 0

## Power of the power

Task 17: Fill in the correct values. correct: 0wrong: 0 Task 18: Simplify the terms and enter the solution.

 a) (43)2 = 4 = b) (24)3 = 2 = c) (72)2 = 7 = d) (102)4 = 10 = e) (52)-2 = 5 = f) (0,1-3)2 = 0,1 = g) (22 · 33 )2 = 2 · = h) (22 · 42)3 = =

Experiments: 0

Task 19: Click on whether the red term summarized 3x3, 3x4, or 3x5 results. Sixteen terms are to be assigned. correct: 0 | wrong: 0 Task 20: Fill in the missing exponents in the multiplication wall.

 a – b a – b a – b a – b a – b a – b a – b a2 b2 a – b

Try: 0 Task 21: Complete the simplified terms correctly. correct: 0wrong: 0 Exercise 22: Fill in the missing terms correctly. correct: 0wrong: 0 Task 23: Fill in the missing values.

 a) pm – p0 – pn = p b) yx + 2 – y – yx – 2 – yx = y c) am – bn – a – b2n = a – b d) (t7 – t2) : (t – t3)= t e) 4-3 : 4-5 = 4

Attempts: 0 Task 24: Enter the missing values. Calculate without a calculator.

 a) 23 · 33 · 43 · 24-2 = b) (76)2 – (73)4 = c) d) (47 · 27)2 : 812 =

Attempts: 0

## Negative Exponents

Exercise 25: Powers can also have negative exponents. Continue the power series and click on the value that appears in the red frame.

 Power 24 23 22 21 20 Number 16 8 4 2 1 Ratio :2 :2 :2 :2 :2 :2 :2 :2

2-4 2-3 2-2 2-1 Attempts: 0 Info: If powers have a negative integer as an exponent, then they can also be written as follows:

 2-2 = 1 = 1 = 0,25 22 4

Task 26: Enter the missing power in the denominator.

 a) 2-6 = 1 b) 3-3 = 1 c) 4-2 = 1

 d) 6-8 = 1 e) 5-2 = 1 f) 8-7 = 1

Attempts: 0 Task 27: Fill in the missing values.

 a) = 1 b) =

 c) 1 = d) 1 = 2 2

correct: 0wrong: 0 Task 28: Fill in the missing denominators and enter the truncated fraction.

 a) 8 · 2-4 = 8 =
 b) 6 · 3-2 = 6 =
 c) 10 · 4-1 = 10 =
 d) 15 · 5-2 = 15 =
 e) 75 · 10-2 = 75 =
 f) 7 · 21-1 = 7 =

Attempts: 0 Task 29: Fill in the missing denominators and enter the correct decimal numbers. Attempts: 0 Task 30: Click on whether the red power value is positive or negative. Eight values have to be assigned. correct: 0 | wrong: 0 Task 31: Complete the mnemonics correctly.

• If the base of a power is positive, then the power value is .
• If the base of a power is negative and the exponent is an even number, then the power value is .
• If the base of a power is negative and the exponent is an odd number, then the power value is .

Attempts: 0 Task 32: Click the powers in the correct order of magnitude. (-4)2 23 112 -(53) (-7)3 (-33) Attempts: 0 Task 33: Click the powers in the correct order of magnitude. (-3)2 (-5)1 -(2)5 (-3)3 (-5)2 (-2)4 Attempts: 0 Task 34: Click whether the result of the red term is positive or negative, if x is a natural number (1, 2, 3 …). Ten values are to be assigned. correct: 0 | wrong: 0 Exponent Examples Math.